NCERT Class 11 Mathematics — Permutations and Combinations, Exercise 6.3. All 11 questions solved.

A permutation is an arrangement, so order matters. The count of arrangements of $r$ objects chosen from $n$ distinct objects is

$$^n\mathrm{P}_r = \frac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1)$$

When some objects are identical, arrangements that merely swap identical objects are the same arrangement, so divide them out. For $n$ objects of which $p$ are alike of one kind, $q$ alike of another, and so on:

$$\frac{n!}{p!\,q!\,\cdots}$$

Two standard devices appear repeatedly:

• Objects that must stay together: glue them into a single block, arrange the blocks, then arrange within the block.
• Objects that must stay apart: count all arrangements and subtract those where they are together.

Key insight. Fixed positions cost nothing and shrink the problem. In question 11(i), $\mathrm P$ and $\mathrm S$ are pinned to the ends, so the count is just the arrangements of the ten letters in between. Recognising which positions are already decided, and removing them from the pool, is what turns a twelve-letter problem into a ten-letter one.

Question 1

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

Solution. Nine digits available and none is $0$, so every arrangement is a valid three-digit number:

$$^9\mathrm{P}_3 = 9 \times 8 \times 7 = 504$$

Question 2

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

Solution. Here the pool is all ten digits $0$ to $9$, and the leading digit cannot be $0$ — otherwise the number is not four-digit. So fill the thousands slot first, where the constraint lives.

• Thousands: any of $1$ to $9$ — $9$ choices.
• Hundreds: any digit except the one used — but $0$ is now allowed, so $9$ again.
• Tens: $8$ choices. Units: $7$ choices.

$$9 \times 9 \times 8 \times 7 = 4536$$

The two nines are not a typo: the first excludes $0$, the second excludes only the digit already used.

Question 3

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

Solution. Even means the units digit is $2$, $4$ or $6$ — three choices, and this is the constrained slot, so fill it first. The remaining two slots then draw from the five unused digits:

$$3 \times 5 \times 4 = 60$$

Question 4

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. Total: four slots from five digits, none of which is $0$:

$$^5\mathrm{P}_4 = 5 \times 4 \times 3 \times 2 = 120$$

Even: the units digit must be $2$ or $4$ — two choices — and the remaining three slots take the four unused digits:

$$2 \times (4 \times 3 \times 2) = 2 \times 24 = 48$$

Question 5

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

Solution. The two posts are different, so order matters — appointing A as chairman and B as vice chairman is not the same as the reverse. This is a permutation, not a selection:

$$^8\mathrm{P}_2 = 8 \times 7 = 56$$

Question 6

Find $n$ if $^{n-1}\mathrm{P}_3 : {}^n\mathrm{P}_4 = 1 : 9$.

Solution. Write the ratio as an equation:

$$\frac{^{n-1}\mathrm{P}_3}{^n\mathrm{P}_4} = \frac19 \quad\Longrightarrow\quad 9 \cdot {}^{n-1}\mathrm{P}_3 = {}^n\mathrm{P}_4$$

Now expand both sides as falling products:

$$^{n-1}\mathrm{P}_3 = (n-1)(n-2)(n-3), \qquad ^n\mathrm{P}_4 = n(n-1)(n-2)(n-3)$$

The right side is exactly $n$ times the left, so

$$9\,(n-1)(n-2)(n-3) = n\,(n-1)(n-2)(n-3)$$

The common factor $(n-1)(n-2)(n-3)$ is non-zero (for the permutations to be defined we need $n \ge 5$), so dividing gives

$$n = 9$$

Check: $^8\mathrm{P}_3 = 336$ and $^9\mathrm{P}_4 = 3024 = 9 \times 336$ ✓

Question 7

Find $r$ if (i) $^5\mathrm{P}_r = 2 \cdot {}^6\mathrm{P}_{r-1}$    (ii) $^5\mathrm{P}_r = {}^6\mathrm{P}_{r-1}$

Solution. Convert to factorials:

$$^5\mathrm{P}_r = \frac{5!}{(5-r)!}, \qquad ^6\mathrm{P}_{r-1} = \frac{6!}{(6-r+1)!} = \frac{6!}{(7-r)!}$$

(i)

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

Since $6! = 6 \times 5!$, cancel $5!$ from both sides:

$$\frac{1}{(5-r)!} = \frac{12}{(7-r)!}$$

Now $(7-r)! = (7-r)(6-r)(5-r)!$, so cancelling $(5-r)!$:

$$(7-r)(6-r) = 12$$

Expanding: $42 – 13r + r^2 = 12$, that is

$$r^2 – 13r + 30 = 0 \quad\Longrightarrow\quad (r-3)(r-10) = 0$$

so $r = 3$ or $r = 10$. But $^5\mathrm{P}_r$ requires $r \le 5$, so $r = 10$ is rejected:

$$r = 3$$

(ii) The same working with $2$ replaced by $1$:

$$\frac{1}{(5-r)!} = \frac{6}{(7-r)!} \quad\Longrightarrow\quad (7-r)(6-r) = 6$$

$$r^2 – 13r + 36 = 0 \quad\Longrightarrow\quad (r-4)(r-9) = 0$$

Rejecting $r = 9$ as $r \le 5$:

$$r = 4$$

Question 8

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

Solution. EQUATION has $8$ letters — E, Q, U, A, T, I, O, N — and all are different, so no division is needed:

$$8! = 40320$$

Question 9

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 the first letter is a vowel?

Solution. MONDAY has $6$ distinct letters: M, O, N, D, A, Y. Its vowels are O and A.

(i) Choose and arrange $4$ of the $6$:

$$^6\mathrm{P}_4 = 6 \times 5 \times 4 \times 3 = 360$$

(ii) All six:

$$6! = 720$$

(iii) The first slot must hold a vowel — $2$ choices — and the remaining five letters fill the rest in any order:

$$2 \times 5! = 2 \times 120 = 240$$

Question 10

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

Solution. MISSISSIPPI has $11$ letters: M ($1$), I ($4$), S ($4$), P ($2$).

Total distinct permutations:

$$\frac{11!}{4!\,4!\,2!} = \frac{39916800}{24 \times 24 \times 2} = \frac{39916800}{1152} = 34650$$

Those with all four I’s together: glue the four I’s into a single block. The block plus the remaining seven letters — M, S, S, S, S, P, P — makes $8$ units, of which the four S’s are alike and the two P’s are alike:

$$\frac{8!}{4!\,2!} = \frac{40320}{48} = 840$$

The I’s inside the block are identical, so there is no extra factor for arranging them — a point where an unnecessary $\times 4!$ often creeps in.

Not together is the complement:

$$34650 – 840 = 33810$$

Question 11

In how many ways can the letters of the word PERMUTATIONS be arranged if

(i) the words start with P and end with S, (ii) the vowels are all together, (iii) there are always $4$ letters between P and S?

Solution. PERMUTATIONS has $12$ letters, and T occurs twice; every other letter appears once. The vowels are E, U, A, I, O — five of them, all distinct. The consonants are P, R, M, T, T, N, S — seven, with T repeated.

(i) P and S are pinned to the two ends, so only the remaining $10$ letters are free. Among those ten, T still appears twice:

$$\frac{10!}{2!} = \frac{3628800}{2} = 1814400$$

(ii) Treat the five vowels as a single block. That block together with the seven consonants gives $8$ units, and the two T’s among the consonants are still alike:

$$\frac{8!}{2!} = \frac{40320}{2} = 20160 \text{ arrangements of the units}$$

Within the block, the five distinct vowels can be ordered in $5! = 120$ ways.

$$20160 \times 120 = 2419200$$

The division by $2!$ is easy to lose here — the block construction does not remove the repeated T.

(iii) “Always $4$ letters between P and S” means P and S occupy positions $i$ and $i+5$. With $12$ positions, $i$ can run from $1$ to $7$ — that is $7$ position pairs — and for each pair P and S can be swapped, giving $2$ orders:

$$7 \times 2 = 14 \text{ ways to place P and S}$$

The remaining $10$ letters (still containing two T’s) fill the other positions:

$$\frac{10!}{2!} = 1814400$$

$$14 \times 1814400 = 25401600$$

Common mistakes

• Question 2, allowing $0$ in the leading position. $^{10}\mathrm{P}_4 = 5040$ counts strings like $0123$, which are not four-digit numbers. The correct count is $9 \times 9 \times 8 \times 7$.
• Question 5, using a combination. Chairman and vice chairman are distinct roles, so the answer is $^8\mathrm{P}_2 = 56$, not $^8\mathrm{C}_2 = 28$. Ask whether swapping the two chosen people changes the outcome — here it does.
• Question 7, keeping the second root. Both parts produce a quadratic with two roots, and in each case the larger one exceeds $5$. Since $^5\mathrm{P}_r$ needs $r \le 5$, it must be discarded — stating that rejection is part of the answer.
• Question 10, multiplying the block by $4!$. The four I’s are identical, so their internal order does not create new arrangements. Contrast question 11(ii), where the five vowels are distinct and $5!$ genuinely applies.
• Question 10, forgetting to subtract. The question asks for arrangements where the I’s are not together, so the block count must be taken away from the total.
• Question 11, forgetting the repeated T. Every count in this question carries a $\div 2!$. Dropping it doubles all three answers.
• Question 11(iii), miscounting the positions. With exactly four letters between them, P and S sit five apart, so the first can start at any of positions $1$ to $7$ — seven pairs, not six or eight — and each pair admits two orders.

Practise next

• Exercise 6.4 — combinations, where order stops mattering and every permutation count is divided by $r!$.
• Miscellaneous Exercise on Chapter 6 — problems that mix selection with arrangement, including the dictionary-order question on EXAMINATION.