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

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 $P(n, r) = \frac{n!}{(n-r)!}$.

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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.

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

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

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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 = $5040 – 504 = 4536$.)

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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).$$\text{Number of ways} = 3 \times P(5, 2)$$$$3 \times (5 \times 4) = 3 \times 20 = \mathbf{60}$$

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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).$$\text{Even numbers} = 2 \times P(4, 3)$$$$2 \times (4 \times 3 \times 2) = 2 \times 24 = \mathbf{48}$$

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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.

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

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

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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:

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

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

$$\mathbf{n = 9}$$

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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 $\mathbf{r = 4}$.

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

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

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

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

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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)$$\text{Number of ways} = 2 \times 5!$$$$2 \times 120 = \mathbf{240}$$

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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:

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

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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.$$\text{Arrangement of } 10 \text{ letters} = \frac{10!}{2!}$$$$\frac{10!}{2!} = 1,814,400$$$$\text{Total arrangements} = (\text{Ways to place P and S}) \times (\text{Arrangement of remaining letters})$$$$\text{Total arrangements} = 14 \times \frac{10!}{2!} = 14 \times 1,814,400 = \mathbf{25,401,600}$$