Rbse Solutions for Class 11 maths Chapter 6 Exercise 6.3 | Permutations

Last Updated on November 24, 2025 by Aman Singh

This exercise focuses on permutations, which is the arrangement of objects where the order matters. The formula for the number of permutations of $n$ distinct objects taken $r$ at a time is $P(n, r) = \frac{n!}{(n-r)!}$.

1. 3-Digit Numbers from Digits 1 to 9 (No Repetition)

We have $n=9$ digits and we are choosing and arranging $r=3$ of them.

$$\text{Number of ways} = P(9, 3)$$

$$P(9, 3) = 9 \times 8 \times 7 = \mathbf{504}$$

2. 4-Digit Numbers with No Digit Repeated

The available digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ($n=10$). The number cannot start with 0.

1. First place (Thousands): Cannot be 0. (9 choices: 1 to 9)
2. Remaining places: We have used 1 digit. We need to choose and arrange the remaining 3 digits from the remaining 9 available digits.$$\text{Number of ways} = 9 \times P(9, 3)$$$$9 \times (9 \times 8 \times 7) = 9 \times 504 = \mathbf{4536}$$(Alternatively: Total 4-digit arrangements (starting with 0 allowed) = $P(10, 4) = 5040$. Arrangements starting with 0 = $P(9, 3) = 504$. Total numbers = $5040 – 504 = 4536$.)

3. 3-Digit Even Numbers from Digits 1, 2, 3, 4, 6, 7 (No Repetition)

Total available digits: $n=6$. The even digits are $\{2, 4, 6\}$ (3 choices).

The Units place must be an even digit.

1. Units place (U): 3 choices (2, 4, or 6)
2. Remaining 2 places (H, T): We have used 1 digit. We choose and arrange 2 digits from the remaining 5 digits. ($P(5, 2)$ ways).$$\text{Number of ways} = 3 \times P(5, 2)$$$$3 \times (5 \times 4) = 3 \times 20 = \mathbf{60}$$

4. 4-Digit Numbers from Digits 1, 2, 3, 4, 5 (No Repetition)

Total available digits: $n=5$. We arrange $r=4$ digits.

• Total 4-digit numbers:$$\text{Total ways} = P(5, 4)$$$$P(5, 4) = 5 \times 4 \times 3 \times 2 = \mathbf{120}$$
• How many of these will be even?The even digits are $\{2, 4\}$ (2 choices). The Units place is restricted.
  1. Units place (U): 2 choices (2 or 4)
  2. Remaining 3 places (H, T, Th): We choose and arrange 3 digits from the remaining 4 digits ($P(4, 3)$ ways).$$\text{Even numbers} = 2 \times P(4, 3)$$$$2 \times (4 \times 3 \times 2) = 2 \times 24 = \mathbf{48}$$

5. Choosing Chairman and Vice Chairman from 8 Persons

We have $n=8$ persons and we are arranging $r=2$ positions (Chairman and Vice Chairman). Order matters as the roles are distinct.

$$\text{Number of ways} = P(8, 2)$$

$$P(8, 2) = 8 \times 7 = \mathbf{56}$$

6. Find $n$ if $P(n – 1, 3) : P(n, 4) = 1 : 9$

Write the permutations in terms of factorials:

$$\frac{P(n – 1, 3)}{P(n, 4)} = \frac{1}{9}$$

$$\frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{(n-4)!}} = \frac{1}{9}$$

$$\frac{(n-1)!}{(n-4)!} \times \frac{(n-4)!}{n!} = \frac{1}{9}$$

$$\frac{(n-1)!}{n!} = \frac{1}{9}$$

Since $n! = n \times (n-1)!$, we substitute this into the denominator:

$$\frac{(n-1)!}{n \times (n-1)!} = \frac{1}{9}$$

$$\frac{1}{n} = \frac{1}{9}$$

$$\mathbf{n = 9}$$

7. Find $r$

The formula $P(n, r) = \frac{n!}{(n-r)!}$ is used.

(i) $P(5, r) = 2 \times P(6, r-1)$

$$\frac{5!}{(5 – r)!} = 2 \times \frac{6!}{(6 – (r – 1))!}$$

$$\frac{5!}{(5 – r)!} = 2 \times \frac{6!}{(7 – r)!}$$

Expand $6! = 6 \cdot 5!$ and $(7 – r)! = (7 – r)(6 – r)(5 – r)!$:

$$\frac{5!}{(5 – r)!} = 2 \times \frac{6 \cdot 5!}{(7 – r)(6 – r)(5 – r)!}$$

Cancel $5!$ and $(5 – r)!$ from both sides:

$$1 = \frac{12}{(7 – r)(6 – r)}$$

$$(7 – r)(6 – r) = 12$$

$$42 – 7r – 6r + r^2 = 12$$

$$r^2 – 13r + 42 – 12 = 0$$

$$r^2 – 13r + 30 = 0$$

Factor the quadratic equation:

$$(r – 3)(r – 10) = 0$$

$$r = 3 \quad \text{or} \quad r = 10$$

Since the constraint for $P(5, r)$ is $r \le 5$, we must have $\mathbf{r = 3}$.

(ii) $P(5, r) = P(6, r-1)$

$$\frac{5!}{(5 – r)!} = \frac{6!}{(6 – (r – 1))!}$$

$$\frac{5!}{(5 – r)!} = \frac{6 \cdot 5!}{(7 – r)!}$$

$$\frac{1}{(5 – r)!} = \frac{6}{(7 – r)(6 – r)(5 – r)!}$$

$$1 = \frac{6}{(7 – r)(6 – r)}$$

$$(7 – r)(6 – r) = 6$$

$$42 – 13r + r^2 = 6$$

$$r^2 – 13r + 36 = 0$$

Factor the quadratic equation:

$$(r – 4)(r – 9) = 0$$

$$r = 4 \quad \text{or} \quad r = 9$$

Since the constraint for $P(5, r)$ is $r \le 5$, we must have $\mathbf{r = 4}$.

8. Words from the letters of EQUATION (Each letter once)

The word EQUATION has $n=8$ distinct letters. We are arranging all 8 letters.

$$\text{Number of words} = 8!$$

$$8! = \mathbf{40320}$$

9. Words from the letters of MONDAY (No Repetition)

The word MONDAY has $n=6$ distinct letters. Vowels are $\{O, A\}$.

(i) 4 letters are used at a time

We are arranging $r=4$ letters from $n=6$ distinct letters.

$$\text{Number of ways} = P(6, 4)$$

$$P(6, 4) = 6 \times 5 \times 4 \times 3 = \mathbf{360}$$

(ii) all letters are used at a time

$$\text{Number of ways} = 6!$$

$$6! = \mathbf{720}$$

(iii) all letters are used but first letter is a vowel

1. First place: Must be a vowel. (2 choices: O or A)
2. Remaining 5 places: We arrange the remaining 5 distinct letters. (5! ways)$$\text{Number of ways} = 2 \times 5!$$$$2 \times 120 = \mathbf{240}$$

10. Distinct Permutations of MISSISSIPPI where the four I’s do not come together

The word MISSISSIPPI has 11 letters.

• M: 1
• I: 4
• S: 4
• P: 2$$\text{Total Permutations} = \frac{11!}{4! 4! 2!} = 34650$$

We use the principle:

$$\text{Arrangements (I’s separate)} = \text{Total Arrangements} – \text{Arrangements (I’s together)}$$

Arrangements where the four I’s are together:

Treat the four I’s as a single block (IIII). We now arrange the 8 items: $\{M, S, S, S, S, P, P, (IIII)\}$.

$$\text{Arrangements (I’s together)} = \frac{8!}{4! 2!}$$

$$\frac{8!}{4! 2!} = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4! \times 2} = \frac{8 \times 7 \times 6 \times 5}{2} = 4 \times 7 \times 6 \times 5 = 840$$

Arrangements where the four I’s do not come together:

$$\text{Result} = 34650 – 840 = \mathbf{33810}$$

11. Permutations of PERMUTATIONS

The word PERMUTATIONS has 12 letters. The letter T is repeated twice.

• T: 2
• All other 10 letters are distinct.$$\text{Total Permutations} = \frac{12!}{2!} = 239,500,800$$Vowels are $\{E, U, A, I, O\}$ (5 vowels).

(i) words start with P and end with S

1. First place: Fixed as P (1 way).
2. Last place: Fixed as S (1 way).
3. Remaining 10 letters: We arrange the remaining 10 letters (E, R, M, U, T, A, T, I, O, N). The letter T is repeated twice.$$\text{Number of ways} = 1 \times \frac{10!}{2!} \times 1$$$$\frac{10!}{2!} = \frac{3,628,800}{2} = \mathbf{1,814,400}$$

(ii) vowels are all together

1. Treat Vowels as a block (V): $V = (E, U, A, I, O)$. The block has $5!$ internal arrangements.
2. Arrange the 8 items: $\{P, R, M, T, T, N, S, (V)\}$. Note that T is repeated twice.$$\text{Arrangement of 8 items} = \frac{8!}{2!}$$$$\text{Number of ways} = (\text{Arrangement of } 8 \text{ items}) \times (\text{Internal arrangement of vowels})$$$$\text{Number of ways} = \frac{8!}{2!} \times 5!$$$$\frac{40320}{2} \times 120 = 20160 \times 120 = \mathbf{2,419,200}$$

(iii) there are always 4 letters between P and S

The positions for P and S must be 4 places apart, e.g., P_ _ _ _ S or S_ _ _ _ P.

1. Position the P and S:
   • Possible starting positions for P (or S): $1, 2, \dots, 7$. (If P is at position 7, S is at 12).
   • Total possible positions for the pair: $12 – 5 = 7$ pairs of positions.
   • For each pair, P and S can be arranged in $2! = 2$ ways.$$\text{Ways to place P and S} = 7 \times 2 = 14$$
2. Arrange the Remaining 10 letters: The remaining 10 letters (E, R, M, U, T, A, T, I, O, N) are arranged in the remaining 10 places. T is repeated twice.$$\text{Arrangement of } 10 \text{ letters} = \frac{10!}{2!}$$$$\frac{10!}{2!} = 1,814,400$$$$\text{Total arrangements} = (\text{Ways to place P and S}) \times (\text{Arrangement of remaining letters})$$$$\text{Total arrangements} = 14 \times \frac{10!}{2!} = 14 \times 1,814,400 = \mathbf{25,401,600}$$