Exercise 6.1 - Question 1

Question: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition of the digits is not allowed?

Solution: There are 5 given digits: $1, 2, 3, 4, 5$. We need to form 3-digit numbers. Thus, there are 3 vacant places to be filled: $_\square\ _\square\ _\square$.

(i) When repetition of the digits is allowed:

• The units place can be filled by any of the 5 digits in $5$ ways.
• The tens place can also be filled by any of the 5 digits in $5$ ways (since repetition is allowed).
• The hundreds place can also be filled by any of the 5 digits in $5$ ways (since repetition is allowed).

• The hundreds place can be filled by any of the 5 digits in $5$ ways.
• The tens place can be filled by any of the remaining 4 digits in $4$ ways.
• The units place can be filled by any of the remaining 3 digits in $3$ ways.

Exercise 6.1 - Question 2

Question: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Solution: Given digits: $1, 2, 3, 4, 5, 6$. Total number of available digits = $6$. We need to form a 3-digit even number where repetition of digits is allowed.

For a number to be even, its units place must be filled by an even digit. The available even digits from the given set are $2, 4,$ and $6$ (total 3 even digits).

• Units place: Can be filled by any of the even digits ($2, 4,$ or $6$) in $3$ ways.
• Tens place: Since repetition is allowed, it can be filled by any of the 6 given digits in $6$ ways.
• Hundreds place: Since repetition is allowed, it can also be filled by any of the 6 given digits in $6$ ways.

Exercise 6.1 - Question 3

Question: How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

Solution: The total number of available letters from the first 10 letters of the English alphabet is $10$. We need to form a 4-letter code such that no letter is repeated.

This can be viewed as filling 4 vacant places using 10 distinct objects without repetition (which is equivalent to permutations):

• 1st letter of the code: Can be chosen in $10$ ways.
• 2nd letter of the code: Can be chosen from the remaining 9 letters in $9$ ways.
• 3rd letter of the code: Can be chosen from the remaining 8 letters in $8$ ways.
• 4th letter of the code: Can be chosen from the remaining 7 letters in $7$ ways.

Exercise 6.1 - Question 4

Question: How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

Solution:
A 5-digit telephone number has 5 blank positions: $\underline{\text{ }} \text{ } \underline{\text{ }} \text{ } \underline{\text{ }} \text{ } \underline{\text{ }} \text{ } \underline{\text{ }}$
It is given that each number starts with $67$. So, the first two places are fixed with digits $6$ and $7$ respectively.
Therefore, the number looks like: $6 \text{ } 7 \text{ } \underline{\text{ }} \text{ } \underline{\text{ }} \text{ } \underline{\text{ }}$
Since no digit can be repeated and the digits $6$ and $7$ are already used, we are left with $10 - 2 = 8$ available digits (0, 1, 2, 3, 4, 5, 8, 9) to fill the remaining 3 places.
The third place can be filled in $8$ ways.
The fourth place can be filled in $7$ ways.
The fifth place can be filled in $6$ ways.
Using the fundamental principle of counting, the total number of 5-digit telephone numbers that can be constructed is:
$$\text{Total numbers} = 8 \times 7 \times 6 = 336$$

Exercise 6.1 - Question 5

Question: A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Solution:
When a coin is tossed, there are $2$ possible outcomes: Head ($H$) or Tail ($T$).
The coin is tossed 3 times.
By the fundamental principle of counting, the total number of possible outcomes is the product of the number of possible outcomes at each of the 3 tosses:
$$\text{Total outcomes} = 2 \times 2 \times 2 = 2^3 = 8$$

Exercise 6.1 - Question 6

Question: Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Solution:
Each signal requires the use of $2$ flags, one below the other. This means we need to fill $2$ vacant places: the upper flag position and the lower flag position.
There are $5$ flags of different colours available.
- The upper flag position can be filled by any of the $5$ flags in $5$ ways.
- After choosing the upper flag, the lower flag position can be filled by any of the remaining $4$ flags in $4$ ways.
Using the fundamental principle of counting, the total number of signals that can be generated is:
$$\text{Total signals} = 5 \times 4 = 20$$

Exercise 6.2 - Question 1

Question: Evaluate (i) $8!$ (ii) $4! - 3!$

Solution:
(i) Evaluate $8!$:
$$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$$
$$8! = 40320$$


(ii) Evaluate $4! - 3!$:
$$4! = 4 \times 3 \times 2 \times 1 = 24$$
$$3! = 3 \times 2 \times 1 = 6$$
$$4! - 3! = 24 - 6 = 18$$

Exercise 6.2 - Question 2

Question: Is $3! + 4! = 7!$?

Solution: To determine whether $3! + 4! = 7!$, let us compute the values of the left-hand side (LHS) and the right-hand side (RHS) independently. First, recall the definition of factorial: $n! = n \times (n-1) \times \dots \times 1$. LHS: $$3! + 4! = (3 \times 2 \times 1) + (4 \times 3 \times 2 \times 1)$$ $$3! = 6$$ $$4! = 24$$ $$\text{LHS} = 6 + 24 = 30$$ RHS: $$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$$ $$7! = 42 \times 120 = 5040$$ Since $\text{LHS} = 30$ and $\text{RHS} = 5040$, we have: $$30 \neq 5040$$ Answer: No, $3! + 4! \neq 7!$.

Exercise 6.2 - Question 3

Question: Compute $\frac{8!}{6! \times 2!}$

Solution: To compute the value of the given expression: $$\frac{8!}{6! \times 2!}$$ We can expand the factorial in the numerator until we reach the largest factorial in the denominator ($6!$) to simplify the expression: $$8! = 8 \times 7 \times 6!$$ Now, substitute this back into the expression: $$\frac{8 \times 7 \times 6!}{6! \times 2!}$$ Cancel out the common term $6!$ from the numerator and the denominator: $$= \frac{8 \times 7}{2!}$$ Evaluate $2!$: $$2! = 2 \times 1 = 2$$ Thus, $$= \frac{8 \times 7}{2} = \frac{56}{2} = 28$$ Answer: $28$

Exercise 6.2 - Question 4

Question: If $\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}$, find $x$.

Solution: Given equation: $$\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}$$ We can expand the factorials in the denominators of the left-hand side to have a common term $6!$: $$7! = 7 \times 6!$$ $$8! = 8 \times 7 \times 6! = 56 \times 6!$$ Substitute these into the equation: $$\frac{1}{6!} + \frac{1}{7 \times 6!} = \frac{x}{56 \times 6!}$$ Factor out $\frac{1}{6!}$ from the left-hand side: $$\frac{1}{6!} \left(1 + \frac{1}{7}\right) = \frac{x}{56 \times 6!}$$ Simplify the term inside the parentheses: $$1 + \frac{1}{7} = \frac{8}{7}$$ So the equation becomes: $$\frac{1}{6!} \times \frac{8}{7} = \frac{x}{56 \times 6!}$$ Cancel out $\frac{1}{6!}$ from both sides: $$\frac{8}{7} = \frac{x}{56}$$ Now, solve for $x$ by cross-multiplication: $$x = \frac{8 \times 56}{7}$$ $$x = 8 \times 8 = 64$$ Answer: $x = 64$

Exercise 6.2 - Question 5

Question: Evaluate $\frac{n!}{(n-r)!}$, when
(i) $n = 6, r = 2$
(ii) $n = 9, r = 5$

Solution: We need to evaluate the expression $\frac{n!}{(n-r)!}$ for the given values of $n$ and $r$. **(i) When $n = 6, r = 2$:** Substitute the values into the expression: $$\frac{6!}{(6-2)!} = \frac{6!}{4!}$$ Expand $6!$ until $4!$: $$\frac{6 \times 5 \times 4!}{4!}$$ Cancel out $4!$: $$= 6 \times 5 = 30$$ **(ii) When $n = 9, r = 5$:** Substitute the values into the expression: $$\frac{9!}{(9-5)!} = \frac{9!}{4!}$$ Expand $9!$ until $4!$: $$\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!}$$ Cancel out $4!$: $$= 9 \times 8 \times 7 \times 6 \times 5$$ $$= 72 \times 42 \times 5$$ $$= 3024 \times 5 = 15120$$ Answer:
(i) $30$
(ii) $15120$

Exercise 6.3 - Question 1

Question: How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Solution: To form a 3-digit number using the digits from 1 to 9, we need to choose and arrange 3 digits out of 9 available digits.
Since the order matters for forming numbers (e.g., 123 is different from 321) and no digit is repeated, this is given by the number of permutations of 9 things taken 3 at a time, denoted as $^{9}P_{3}$.
The formula for permutations is: $${}^{n}P_{r} = \frac{n!}{(n-r)!}$$
Substitute $n = 9$ and $r = 3$: $${}^{9}P_{3} = \frac{9!}{(9-3)!} = \frac{9!}{6!} = \frac{9 \times 8 \times 7 \times 6!}{6!} = 9 \times 8 \times 7 = 504$$
Answer: 504

Exercise 6.3 - Question 2

Question: How many 4-digit numbers are there with no digit repeated?

Solution:
The available digits for forming any number are $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$, which gives a total of 10 digits.
A 4-digit number cannot have $0$ in the thousands place (the first position).
1. The thousands place can be filled by any of the 9 non-zero digits ($1, 2, 3, 4, 5, 6, 7, 8, 9$). Number of ways = $9$.
2. The hundreds place can be filled by any of the remaining 9 digits (including $0$, since one digit has been used in the thousands place). Number of ways = $9$.
3. The tens place can be filled by any of the remaining 8 digits. Number of ways = $8$.
4. The units place can be filled by any of the remaining 7 digits. Number of ways = $7$.
By the fundamental principle of counting, the total number of 4-digit numbers with no digit repeated is: $$\text{Total numbers} = 9 \times 9 \times 8 \times 7 = 4536$$
Answer: 4536

Exercise 6.3 - Question 3

Question: How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Solution:
The available digits are $\{1, 2, 3, 4, 6, 7\}$, which gives a total of 6 digits. We need to form 3-digit even numbers without repetition.
An even number must end with an even digit. The even digits available in the given set are $\{2, 4, 6\}$.
1. Units place: Can be filled by any of the 3 even digits ($2, 4, \text{ or } 6$). Number of ways = $3$.
2. Tens and Hundreds places: Out of the 6 total digits, 1 digit has been used in the units place. We are left with 5 digits to fill the remaining 2 places. The number of ways to arrange 2 digits out of 5 is: $${}^{5}P_{2} = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20$$
By the fundamental principle of counting, the total number of 3-digit even numbers is: $$\text{Total} = 3 \times 20 = 60$$
Answer: 60

Exercise 6.3 - Question 4

Question: Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

Solution:
The available digits are $\{1, 2, 3, 4, 5\}$, giving a total of $n = 5$ digits. No digit is repeated.

Part 1: Total 4-digit numbers
We need to arrange 4 digits out of 5. This is given by: $${}^{5}P_{4} = \frac{5!}{(5-4)!} = \frac{5!}{1!} = 5 \times 4 \times 3 \times 2 = 120$$

Part 2: Total even 4-digit numbers
For a number to be even, its units place must be filled with an even digit. The even digits available are $\{2, 4\}$, which gives 2 choices for the units place.
1. Units place: Can be filled in $2$ ways.
2. Remaining 3 places (thousands, hundreds, tens): Out of 5 digits, 1 has been used in the units place, leaving 4 digits to fill the remaining 3 places. The number of ways to arrange 3 digits out of 4 is: $${}^{4}P_{3} = \frac{4!}{(4-3)!} = \frac{4!}{1!} = 4 \times 3 \times 2 = 24$$
By the fundamental principle of counting, the total number of even 4-digit numbers is: $$\text{Total even numbers} = 2 \times 24 = 48$$
Answer: 120 (total 4-digit numbers), 48 (even 4-digit numbers)

Exercise 6.3 - Question 5

Question: From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?

Solution:
Total number of persons in the committee, $n = 8$.
We need to choose 2 positions: a chairman and a vice-chairman, such that one person cannot hold more than one position. This means the order of selection matters (selecting person A as chairman and person B as vice-chairman is different from selecting person B as chairman and person A as vice-chairman).
Thus, this is a problem of permutations of $n$ objects taken $r$ at a time, where $n = 8$ and $r = 2$.
The number of ways is given by the formula for permutations: $$^nP_r = \frac{n!}{(n-r)!}$$ Substitute $n = 8$ and $r = 2$: $$^8P_2 = \frac{8!}{(8-2)!} = \frac{8!}{6!} = \frac{8 \times 7 \times 6!}{6!} = 8 \times 7 = 56$$
Answer: 56

Exercise 6.3 - Question 6

Question: Find $n$ if $^{n-1}P_3 : ^nP_4 = 1 : 9$.

Solution:
Given equation: $$^{n-1}P_3 : ^nP_4 = 1 : 9$$ Using the formula for permutations $^{n}P_r = \frac{n!}{(n-r)!}$, we write: $$\frac{^{n-1}P_3}{^nP_4} = \frac{1}{9}$$ $$\frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{(n-4)!}} = \frac{1}{9}$$ $$\frac{\frac{(n-1)!}{(n-4)!}}{\frac{n!}{(n-4)!}} = \frac{1}{9}$$ Multiply by the reciprocal of the denominator: $$\frac{(n-1)!}{(n-4)!} \times \frac{(n-4)!}{n!} = \frac{1}{9}$$ $$\frac{(n-1)!}{n!} = \frac{1}{9}$$ Since $n! = n \times (n-1)!$, we can substitute it into the denominator: $$\frac{(n-1)!}{n \times (n-1)!} = \frac{1}{9}$$ Cancel out $(n-1)!$ from both numerator and denominator: $$\frac{1}{n} = \frac{1}{9}$$ Thus, $$n = 9$$
Answer: $n = 9$

Exercise 6.3 - Question 7

Question: Find $r$ if
(i) $^5P_r = 2 \cdot {^6P_{r-1}}$
(ii) $^5P_r = ^6P_{r-1}$

Solution:
Part (i): $^5P_r = 2 \cdot {^6P_{r-1}}$
Using the formula $^{n}P_r = \frac{n!}{(n-r)!}$: $$\frac{5!}{(5-r)!} = 2 \cdot \frac{6!}{(6-(r-1))!}$$ $$\frac{5!}{(5-r)!} = 2 \cdot \frac{6!}{(7-r)!}$$ Expand $6!$ as $6 \times 5!$ and $(7-r)!$ as $(7-r)(6-r)(5-r)!$: $$\frac{5!}{(5-r)!} = 2 \cdot \frac{6 \times 5!}{(7-r)(6-r)(5-r)!}$$ Divide both sides by $5!$ and $(5-r)!$: $$1 = \frac{12}{(7-r)(6-r)}$$ $$(7-r)(6-r) = 12$$ $$42 - 7r - 6r + r^2 = 12$$ $$r^2 - 13r + 42 - 12 = 0$$ $$r^2 - 13r + 30 = 0$$ Factorize the quadratic equation: $$(r - 10)(r - 3) = 0$$ So, $r = 10$ or $r = 3$. Since $r \le n$ (here $n=5$), $r$ cannot be 10. Therefore, **$r = 3$**.

Part (ii): $^5P_r = ^6P_{r-1}$
Using the formula: $$\frac{5!}{(5-r)!} = \frac{6!}{(6-(r-1))!}$$ $$\frac{5!}{(5-r)!} = \frac{6!}{(7-r)!}$$ Expand $6!$ and $(7-r)!$: $$\frac{5!}{(5-r)!} = \frac{6 \times 5!}{(7-r)(6-r)(5-r)!}$$ Divide both sides by $5!$ and $(5-r)!$: $$1 = \frac{6}{(7-r)(6-r)}$$ $$(7-r)(6-r) = 6$$ $$42 - 13r + r^2 = 6$$ $$r^2 - 13r + 36 = 0$$ Factorize: $$(r - 9)(r - 4) = 0$$ So, $r = 9$ or $r = 4$. Since $r \le 5$, $r$ cannot be 9. Therefore, **$r = 4$**.
Answer: (i) $r = 3$, (ii) $r = 4$

Exercise 6.3 - Question 8

Question: How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

Solution:
Let us count the number of letters in the word EQUATION:
E, Q, U, A, T, I, O, N
Total number of letters, $n = 8$.
All the 8 letters are distinct (no letter is repeated).
We need to form words using all the letters, taking all 8 letters at a time. The number of ways to arrange $n$ distinct objects taken all at a time is given by $n!$.
Thus, the total number of words that can be formed is: $$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$$
Answer: 40320

Exercise 6.3 - Question 9

Question: How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. (i) 4 letters are used at a time, (ii) all letters are used at a time, (iii) all letters are used but first letter is a vowel?

Solution: The given word is **MONDAY**, which contains $6$ distinct letters: M, O, N, D, A, Y. (i) **4 letters are used at a time:** The number of ways of arranging $n$ distinct objects taken $r$ at a time is given by the permutation formula: $$^nP_r = \frac{n!}{(n-r)!}$$ Here, $n = 6$ and $r = 4$. $$^6P_4 = \frac{6!}{(6-4)!} = \frac{6!}{2!} = \frac{6 \times 5 \times 4 \times 3 \times 2!}{2!} = 6 \times 5 \times 4 \times 3 = 360$$ **Answer:** $360$ (ii) **All letters are used at a time:** Here, $n = 6$ and all $6$ letters are used ($r = 6$). $$^6P_6 = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$$ **Answer:** $720$ (iii) **All letters are used but the first letter is a vowel:** The vowels in the word MONDAY are **O** and **A** (total $2$ vowels). - The first position can be filled by any of the $2$ vowels in $^2P_1 = 2$ ways. - Since one vowel is used at the first position and no repetition is allowed, the remaining $5$ positions can be filled using the remaining $5$ letters in $^5P_5 = 5! = 120$ ways. By the fundamental principle of counting, the total number of words is: $$\text{Total words} = 2 \times 5! = 2 \times 120 = 240$$ **Answer:** $240$

Exercise 6.3 - Question 10

Question: In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together?

Solution: The given word is **MISSISSIPPI**. Total number of letters = $11$. The frequency of each letter is: - M: $1$ - I: $4$ - S: $4$ - P: $2$ The total number of distinct permutations of all letters is given by: $$\text{Total permutations} = \frac{11!}{1! \times 4! \times 4! \times 2!}$$ $$\text{Total permutations} = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}{1 \times (4!) \times (4 \times 3 \times 2 \times 1) \times (2 \times 1)}$$ $$= \frac{39,916,800}{24 \times 2} = \frac{39,916,800}{576} = 34,650$$ Next, we find the number of permutations in which **all four I's come together**. Treat the four I's as a single entity/block $(IIII)$. The remaining letters are: M, S, S, S, S, P, P (total $7$ other letters + $1$ block = $8$ entities). Within these $8$ entities, the repetitions are: - S: $4$ times - P: $2$ times The number of arrangements where all four I's are together is: $$\text{Permutations (I's together)} = \frac{8!}{4! \times 2!}$$ $$= \frac{8 \times 7 \times 6 \times 5 \times 4!}{4! \times (2 \times 1)} = \frac{1680}{2} = 840$$ Therefore, the number of permutations in which the four I's **do not** come together is: $$\text{Required permutations} = (\text{Total permutations}) - (\text{Permutations with I's together})$$ $$= 34,650 - 840 = 33,810$$ **Answer:** $33,810$

Exercise 6.3 - Question 11

Question: In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S?

Solution: The word **PERMUTATIONS** has $12$ letters with the following frequencies: - P: $1$ - E: $1$ - R: $1$ - M: $1$ - U: $1$ - T: $2$ - A: $1$ - I: $1$ - O: $1$ - N: $1$ - S: $1$ (i) **Words start with P and end with S:** - Fix **P** at the first position and **S** at the last position ($12^{\text{th}}$ position). - Remaining $10$ positions are to be filled by the remaining $10$ letters: E, R, M, U, T, A, I, O, N, T (note that T repeats twice). $$\text{Number of arrangements} = \frac{10!}{2!} = \frac{3,628,800}{2} = 1,814,400$$ **Answer:** $1,814,400$ (ii) **Vowels are all together:** The vowels in PERMUTATIONS are **E, U, A, I, O** ($5$ vowels, all distinct). Treat these $5$ vowels as a single entity/block. Along with the remaining $7$ consonants (P, R, M, T, T, N, S - total $7$ consonants), we have $1 + 7 = 8$ entities to arrange. The repetitions among these $8$ entities are: - T: $2$ The number of ways to arrange these $8$ entities is: $$\frac{8!}{2!} = \frac{40,320}{2} = 20,160$$ The $5$ vowels inside the block can be arranged among themselves in $5!$ ways: $$5! = 120$$ Total arrangements = $20,160 \times 120 = 2,419,200$ **Answer:** $2,419,200$ (iii) **There are always 4 letters between P and S:** Let the total 12 positions be numbered $1$ to $12$. The pairs of positions $(i, j)$ such that there are exactly $4$ letters between them (i.e., $|i - j| = 5$) are: - $(1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11), (7, 12)$ — $7$ pairs of positions. For each pair, P and S can be placed in $2$ ways (either P at $i$ and S at $j$, or S at $i$ and P at $j$). So, the number of ways to place P and S = $7 \times 2 = 14$ ways. The remaining $10$ positions can be filled by the remaining $10$ letters (which contain T twice, all others distinct): $$\text{Number of ways to arrange remaining 10 letters} = \frac{10!}{2!} = \frac{3,628,800}{2} = 1,814,400$$ Total arrangements = $14 \times 1,814,400 = 25,401,600$. **Answer:** $25,401,600$

Exercise 6.4 - Question 1

Question: If $^nC_8 = ^nC_2$, find $^nC_2$.

Solution: We are given that: $$^nC_8 = ^nC_2$$ We know the standard property of combinations: $$^nC_x = ^nC_y \implies x = y \text{ or } x + y = n$$ Since $8 \neq 2$, we must have: $$8 + 2 = n \implies n = 10$$ Now, we need to find the value of $^nC_2$ for $n = 10$: $$^{10}C_2 = \frac{10!}{2!(10-2)!} = \frac{10!}{2! \times 8!}$$ $$= \frac{10 \times 9 \times 8!}{2 \times 1 \times 8!} = \frac{90}{2} = 45$$ **Answer:** $45$

Exercise 6.4 - Question 2

Question: Determine $n$ if
(i) $^{2n}C_3 : ^nC_3 = 12 : 1$
(ii) $^{2n}C_3 : ^nC_3 = 11 : 1$

Solution: We know that the formula for combinations is given by: $$^nC_r = \frac{n!}{r!(n-r)!}$$ Part (i): $$\frac{^{2n}C_3}{^nC_3} = \frac{12}{1}$$ Using the combination formula: $$\frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}} = 12$$ $$\frac{(2n)!}{(2n-3)!} \times \frac{(n-3)!}{n!} = 12$$ Expanding the factorials: $$\frac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!} \times \frac{(n-3)!}{n(n-1)(n-2)(n-3)!} = 12$$ $$\frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)} = 12$$ Taking $2$ common from $(2n-2)$: $$\frac{(2n)(2n-1) \cdot 2(n-1)}{n(n-1)(n-2)} = 12$$ $$\frac{4(2n-1)}{n-2} = 12$$ Divide both sides by $4$: $$\frac{2n-1}{n-2} = 3$$ Cross-multiplying: $$2n - 1 = 3(n - 2)$$ $$2n - 1 = 3n - 6$$ $$n = 5$$
Part (ii): $$\frac{^{2n}C_3}{^nC_3} = \frac{11}{1}$$ Using the expanded form derived in part (i): $$\frac{4(2n-1)}{n-2} = 11$$ Cross-multiplying: $$4(2n - 1) = 11(n - 2)$$ $$8n - 4 = 11n - 22$$ $$3n = 18$$ $$n = 6$$ Final Answer:
(i) $n = 5$
(ii) $n = 6$

Exercise 6.4 - Question 3

Question: How many chords can be drawn through 21 points on a circle?

Solution: To draw a single chord on a circle, we need to choose exactly $2$ distinct points on the circumference of the circle and join them. Given total number of points $n = 21$. The number of ways to choose $2$ points out of $21$ points is given by the combination formula $^{n}C_r$: $$^{21}C_2 = \frac{21!}{2!(21-2)!}$$ $$^{21}C_2 = \frac{21!}{2! \cdot 19!}$$ $$^{21}C_2 = \frac{21 \times 20 \times 19!}{2 \times 1 \times 19!}$$ $$^{21}C_2 = \frac{21 \times 20}{2} = 21 \times 10 = 210$$

Exercise 6.4 - Question 4

Question: In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Solution: We need to select: - $3$ boys out of $5$ available boys, which can be done in $^{5}C_3$ ways. - $3$ girls out of $4$ available girls, which can be done in $^{4}C_3$ ways. Using the fundamental principle of counting, the total number of ways to select the team is the product of these two combinations: $$\text{Total ways} = (^{5}C_3) \times (^{4}C_3)$$ Calculating $^{5}C_3$: $$^{5}C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3!}{3! \times 2!} = \frac{5 \times 4}{2 \times 1} = 10$$ Calculating $^{4}C_3$: $$^{4}C_3 = \frac{4!}{3!(4-3)!} = \frac{4 \times 3!}{3! \times 1!} = 4$$ Multiplying the results: $$\text{Total ways} = 10 \times 4 = 40$$ Final Answer:
The team can be selected in $40$ ways.

Exercise 6.4 - Question 5

Question: Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Solution: We are given: - Total red balls = $6$, and we need to select $3$ red balls. - Total white balls = $5$, and we need to select $3$ white balls. - Total blue balls = $5$, and we need to select $3$ blue balls. The number of ways to make each selection is: 1. Selecting $3$ red balls out of $6$: $$^{6}C_3 = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4 \times 3!}{3! \times (3 \times 2 \times 1)} = \frac{120}{6} = 20$$ 2. Selecting $3$ white balls out of $5$: $$^{5}C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3!}{3! \times (2 \times 1)} = \frac{20}{2} = 10$$ 3. Selecting $3$ blue balls out of $5$: $$^{5}C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3!}{3! \times (2 \times 1)} = \frac{20}{2} = 10$$ By the Fundamental Principle of Counting, the total number of ways to select the 9 balls is the product of the individual combinations: $$\text{Total ways} = (^{6}C_3) \times (^{5}C_3) \times (^{5}C_3)$$ $$\text{Total ways} = 20 \times 10 \times 10 = 2000$$ Final Answer:
The number of ways of selecting the balls is $2000$.

Exercise 6.4 - Question 6

Question: Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Solution:
A standard deck of cards contains $52$ cards, out of which there are $4$ aces and $48$ non-ace cards.

We need to form a combination of $5$ cards such that each combination contains exactly one ace.

Step 1: Select $1$ ace out of the $4$ available aces. The number of ways to do this is given by: $${}^{4}\text{C}_{1} = \frac{4!}{1!(4-1)!} = 4$$ Step 2: Select the remaining $4$ cards out of the $48$ non-ace cards. The number of ways to do this is given by: $${}^{48}\text{C}_{4} = \frac{48!}{4!(48-4)!} = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1} = 194580$$ Step 3: By the Fundamental Principle of Counting, the total number of such 5-card combinations is the product of the number of ways of both selections: $$\text{Total combinations} = {}^{4}\text{C}_{1} \times {}^{48}\text{C}_{4} = 4 \times 194580 = 778320$$ Final Answer:
The number of 5-card combinations is $778320$.

Exercise 6.4 - Question 7

Question: In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

Solution:
Total number of players available = $17$
Number of players who can bowl = $5$
Number of players who cannot bowl (non-bowlers) = $17 - 5 = 12$

We need to select a team of $11$ players such that the team includes exactly $4$ bowlers.

Step 1: Select $4$ bowlers out of the $5$ available bowlers. The number of ways to do this is: $${}^{5}\text{C}_{4} = \frac{5!}{4!(5-4)!} = 5$$ Step 2: Since a team consists of $11$ players in total and $4$ are already chosen as bowlers, the remaining $11 - 4 = 7$ players must be selected from the $12$ non-bowlers. The number of ways to do this is: $${}^{12}\text{C}_{7} = \frac{12!}{7!(12-7)!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792$$ Step 3: By the Fundamental Principle of Counting, the total number of ways to select the cricket team is: $$\text{Total ways} = {}^{5}\text{C}_{4} \times {}^{12}\text{C}_{7} = 5 \times 792 = 3960$$ Answer: $3960$

Exercise 6.4 - Question 8

Question: A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Solution:
Number of black balls in the bag = $5$
Number of red balls in the bag = $6$

We need to select $2$ black balls and $3$ red balls.

Step 1: Select $2$ black balls out of $5$ black balls. The number of ways is: $${}^{5}\text{C}_{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$$ Step 2: Select $3$ red balls out of $6$ red balls. The number of ways is: $${}^{6}\text{C}_{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$ Step 3: By the Fundamental Principle of Counting, the total number of ways to select the balls is: $$\text{Total ways} = {}^{5}\text{C}_{2} \times {}^{6}\text{C}_{3} = 10 \times 20 = 200$$ Answer: $200$

Exercise 6.4 - Question 9

Question: In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Solution:
Total number of available courses = $9$
Number of courses a student needs to choose = $5$
Number of specific courses that are compulsory = $2$

Since $2$ courses are compulsory, they must be included in every student's programme of $5$ courses.

Step 1: The student has already chosen the $2$ compulsory courses out of the $2$ required.

Step 2: The remaining number of courses to be chosen = $5 - 2 = 3$.
The remaining number of available courses to choose from (excluding the $2$ compulsory ones) = $9 - 2 = 7$.

Step 3: We now need to select $3$ courses out of the remaining $7$ courses. The number of ways to do this is given by: $${}^{7}\text{C}_{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$$ Answer: $35$

Miscellaneous Exercise Chapter 6 - Question 1

Question: How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER ?

Solution:
The word DAUGHTER contains 8 distinct letters in total.
Let us first identify the vowels and consonants in the word DAUGHTER:
- Vowels: $A, U, E$ (Total 3 vowels)
- Consonants: $D, G, H, T, R$ (Total 5 consonants)

We need to form words containing 2 vowels and 3 consonants.
Step 1: Select 2 vowels out of the 3 available vowels.
The number of ways to choose 2 vowels is given by combination:
$$^{3}\text{C}_{2} = \frac{3!}{2!(3-2)!} = 3$$

Step 2: Select 3 consonants out of the 5 available consonants.
The number of ways to choose 3 consonants is:
$$^{5}\text{C}_{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$$

Step 3: Total number of ways to select the group of 5 letters (2 vowels and 3 consonants):
$$\text{Total selections} = ^{3}\text{C}_{2} \times ^{5}\text{C}_{3} = 3 \times 10 = 30$$

Step 4: Each selected group consists of 5 distinct letters, which can be arranged among themselves in $5!$ ways.
$$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$

Step 5: Total number of words formed is:
$$\text{Total words} = 30 \times 120 = 3600$$
Answer: $3600$

Miscellaneous Exercise Chapter 6 - Question 2

Question: How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?

Solution:
The word EQUATION contains 8 letters in total, with all letters being distinct.
- Vowels: $E, U, A, I, O$ (Total 5 vowels)
- Consonants: $Q, T, N$ (Total 3 consonants)

Step 1: Treat all 5 vowels as a single unit (block) $V$, and all 3 consonants as another single unit (block) $C$.
Now we have 2 major units ($V$ and $C$) which can be arranged among themselves in $2!$ ways:
$$2! = 2 \times 1 = 2$$

Step 2: The 5 vowels within their own block ($E, U, A, I, O$) can be arranged among themselves in $5!$ ways:
$$5! = 120$$

Step 3: The 3 consonants within their own block ($Q, T, N$) can be arranged among themselves in $3!$ ways:
$$3! = 6$$

Step 4: Total number of words formed is the product of all arrangements:
$$\text{Total words} = 2! \times 5! \times 3! = 2 \times 120 \times 6 = 1440$$

Miscellaneous Exercise Chapter 6 - Question 3

Question: A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) exactly 3 girls ?
(ii) atleast 3 girls ?
(iii) atmost 3 girls ?

Solution:
Total boys = $9$, Total girls = $4$, Committee size = $7$.

(i) Exactly 3 girls:
If there are exactly 3 girls in the committee of 7, then the number of boys must be $7 - 3 = 4$.
$$\text{Number of ways} = ^{4}\text{C}_{3} \times ^{9}\text{C}_{4}$$
$$^{4}\text{C}_{3} = 4$$
$$^{9}\text{C}_{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$$
$$\text{Total ways} = 4 \times 126 = 504$$

(ii) Atleast 3 girls:
Atleast 3 girls means the committee can have either 3 girls or 4 girls.
- Case 1: 3 girls and 4 boys $\rightarrow ^{4}\text{C}_{3} \times ^{9}\text{C}_{4} = 504$
- Case 2: 4 girls and 3 boys $\rightarrow ^{4}\text{C}_{4} \times ^{9}\text{C}_{3} = 1 \times \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$
$$\text{Total ways} = 504 + 84 = 588$$

(iii) Atmost 3 girls:
Atmost 3 girls means the committee can have 0, 1, 2, or 3 girls.
- Case 1: 0 girls, 7 boys $\rightarrow ^{4}\text{C}_{0} \times ^{9}\text{C}_{7} = 1 \times 36 = 36$
- Case 2: 1 girl, 6 boys $\rightarrow ^{4}\text{C}_{1} \times ^{9}\text{C}_{6} = 4 \times 84 = 336$
- Case 3: 2 girls, 5 boys $\rightarrow ^{4}\text{C}_{2} \times ^{9}\text{C}_{5} = 6 \times 126 = 756$
- Case 4: 3 girls, 4 boys $\rightarrow ^{4}\text{C}_{3} \times ^{9}\text{C}_{4} = 4 \times 126 = 504$
$$\text{Total ways} = 36 + 336 + 756 + 504 = 1632$$

Miscellaneous Exercise Chapter 6 - Question 4

Question: If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E ?

Solution:
The word EXAMINATION has 11 letters in total, with the following frequencies:
- A occurs 2 times
- I occurs 2 times
- N occurs 2 times
- E occurs 1 time
- X occurs 1 time
- M occurs 1 time
- T occurs 1 time
- O occurs 1 time

We need to find the number of words that appear in the dictionary before the first word starting with 'E'.
Since the dictionary lists words alphabetically, the words appearing *before* any word starting with 'E' must start with the letter that comes alphabetically before 'E'.
The only letter in EXAMINATION that comes alphabetically before 'E' is A.

Step 1: Fix 'A' at the first position.
When 'A' is fixed at the first position, we have remaining 10 letters to arrange:
Remaining letters: $E, X, A, M, I, N, A, T, I, O, N$ (excluding one 'A')
Frequencies of remaining 10 letters:
- A: 1
- E: 1
- I: 2
- M: 1
- N: 2
- O: 1
- T: 1
- X: 1
Total = $1 + 1 + 2 + 1 + 2 + 1 + 1 + 1 = 10$ letters.

Step 2: Calculate the number of arrangements of these 10 letters.
$$\text{Number of words starting with A} = \frac{10!}{1! \times 1! \times 2! \times 1! \times 2! \times 1! \times 1! \times 1!} = \frac{10!}{2! \times 2!}$$
$$10! = 3,628,800$$
$$2! \times 2! = 4$$
$$\text{Number of words} = \frac{3628800}{4} = 907200$$

Miscellaneous Exercise Chapter 6 - Question 5

Question: How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated ?

Solution:
Given digits: $0, 1, 3, 5, 7, 9$ (Total 6 digits).
We need to form 6-digit numbers without repetition that are divisible by 10.
A number is divisible by 10 if and only if its units place digit is $0$.
Therefore, the units place can be filled in only $1$ way (by the digit $0$).
Since no digit is repeated and $0$ is fixed at the units place, the remaining 5 places can be filled by the remaining 5 digits ($1, 3, 5, 7, 9$) in $5!$ ways.
Number of ways $= 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
Since $0$ is fixed at the units place, none of the numbers formed by arranging the remaining 5 non-zero digits in the first 5 places will have $0$ at the starting place. The highest place will be filled by one of $\{1, 3, 5, 7, 9\}$.
Thus, all $120$ numbers are valid 6-digit numbers.

Final Answer: $120$

Miscellaneous Exercise - Question 6

Question: The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet ?

Solution:
Given:
Total vowels available $= 5$
Total consonants available $= 21$
We need to form words containing $2$ different vowels and $2$ different consonants.
Step 1: Selection of letters
Number of ways to choose $2$ different vowels out of $5$ is given by:
$^{5}\text{C}_{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$
Number of ways to choose $2$ different consonants out of $21$ is given by:
$^{21}\text{C}_{2} = \frac{21!}{2!(21-2)!} = \frac{21 \times 20}{2 \times 1} = 210$
Total number of ways to select the $4$ letters (2 vowels and 2 consonants) is:
$\text{Selection ways} = {}^{5}\text{C}_{2} \times {}^{21}\text{C}_{2} = 10 \times 210 = 2100$
Step 2: Arrangement of letters
Each selected group contains $4$ different letters.
The number of ways to arrange $4$ distinct letters among themselves is:
$4! = 4 \times 3 \times 2 \times 1 = 24$
Step 3: Total number of words
Total words = (Ways to select the letters) $\times$ (Ways to arrange the selected letters)
$\text{Total words} = 2100 \times 24 = 50400$
Final Answer: $50400$

Miscellaneous Exercise - Question 7

Question: In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions ?

Solution:
Given:
Total questions in Part I $= 5$
Total questions in Part II $= 7$
Total questions to be attempted $= 8$
Condition: At least 3 questions must be selected from each part.
Let $x$ be the number of questions chosen from Part I and $y$ be the number of questions chosen from Part II.
We have: $x + y = 8$, with $x \ge 3$ and $y \ge 3$.
Possible cases for $(x, y)$:
1. $x = 3, y = 5$ (3 from Part I, 5 from Part II)
2. $x = 4, y = 4$ (4 from Part I, 4 from Part II)
3. $x = 5, y = 3$ (5 from Part I, 3 from Part II)
Now, let's calculate the number of ways for each case:
Case 1: 3 questions from Part I and 5 from Part II
$$\text{Ways}_1 = {}^{5}\text{C}_{3} \times {}^{7}\text{C}_{5} = \frac{5!}{3!2!} \times \frac{7!}{5!2!} = 10 \times 21 = 210$$
Case 2: 4 questions from Part I and 4 from Part II
$$\text{Ways}_2 = {}^{5}\text{C}_{4} \times {}^{7}\text{C}_{4} = \frac{5!}{4!1!} \times \frac{7!}{4!3!} = 5 \times 35 = 175$$
Case 3: 5 questions from Part I and 3 from Part II
$$\text{Ways}_3 = {}^{5}\text{C}_{5} \times {}^{7}\text{C}_{3} = \frac{5!}{5!0!} \times \frac{7!}{3!4!} = 1 \times 35 = 35$$
Total number of ways:
$$\text{Total ways} = \text{Ways}_1 + \text{Ways}_2 + \text{Ways}_3 = 210 + 175 + 35 = 420$$
Final Answer: $420$

Miscellaneous Exercise - Question 8

Question: Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

Solution:
Given:
Total number of cards in a deck $= 52$
Number of kings in a deck $= 4$
Number of non-king cards $= 52 - 4 = 48$
We need to select a combination of $5$ cards such that each selection contains:
1. Exactly $1$ king
2. Exactly $4$ non-king cards (since $5 - 1 = 4$)
Step 1: Select 1 king out of 4
Number of ways to choose $1$ king $= {}^{4}\text{C}_{1}$
Step 2: Select 4 non-king cards out of 48
Number of ways to choose $4$ cards from the remaining $48$ cards $= {}^{48}\text{C}_{4}$
Step 3: Total number of combinations
By the fundamental principle of counting, the total number of valid 5-card combinations is:
$$\text{Total combinations} = {}^{4}\text{C}_{1} \times {}^{48}\text{C}_{4}$$
Final Answer: ${}^{4}\text{C}_{1} \times {}^{48}\text{C}_{4}$

Miscellaneous Exercise - Question 9

Question: It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible ?

Solution:
Total number of persons = $5 \text{ (men)} + 4 \text{ (women)} = 9$ persons.
Total number of available places in a row = $9$.
The positions in the row can be marked as:
1 (odd), 2 (even), 3 (odd), 4 (even), 5 (odd), 6 (even), 7 (odd), 8 (even), 9 (odd).
There are 4 even places (2nd, 4th, 6th, and 8th) and 5 odd places (1st, 3rd, 5th, 7th, and 9th).

Step 1: Arrange the 4 women in the 4 even places.
The number of ways to arrange 4 women in 4 places is given by:
$$^4P_4 = 4! = 4 \times 3 \times 2 \times 1 = 24$$

Step 2: Arrange the 5 men in the remaining 5 odd places.
The number of ways to arrange 5 men in 5 places is given by:
$$^5P_5 = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$

Step 3: Calculate the total number of arrangements by the fundamental principle of counting.
$$\text{Total arrangements} = 4! \times 5! = 24 \times 120 = 2880$$
Answer: $2880$

Exercise 6.4 - Question 10

Question: From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen ?

Solution:
Total number of students in the class = $25$.
Number of students to be chosen for the excursion party = $10$.
There is a specific group of $3$ students who either all join or none join. We can consider two mutually exclusive cases:

Case I: When all 3 students join the excursion party.
Since these 3 students are definitely included, we now need to select the remaining $10 - 3 = 7$ students from the remaining $25 - 3 = 22$ students.
The number of ways to do this is:
$$^{22}C_7 = \frac{22!}{7!(22-7)!} = \frac{22!}{7!15!}$$

Case II: When none of the 3 students join the excursion party.
Since these 3 students are excluded, we need to select all $10$ students from the remaining $25 - 3 = 22$ students.
The number of ways to do this is:
$$^{22}C_{10} = \frac{22!}{10!(22-10)!} = \frac{22!}{10!12!}$$

Total number of ways:
By the addition principle, the total number of ways to choose the excursion party is the sum of the ways from both cases:
$$\text{Total ways} = ^{22}C_7 + ^{22}C_{10}$$
Answer: $^{22}C_7 + ^{22}C_{10}$

Miscellaneous Exercise on Chapter 6 - Question 11

Question: In how many ways can the letters of the word ASSASSINATION be arranged so that all the S's are together ?

Solution:
The given word is ASSASSINATION.
Let us count the total frequency of each letter in the word:

• A: 3
• S: 4
• I: 2
• N: 2
• T: 1
• O: 1