NCERT Class 11 Mathematics — Permutations and Combinations, Miscellaneous Exercise on Chapter 6. All 11 questions solved.

Almost every question here is a two-stage problem: first select, then arrange what was selected. Getting the two stages the right way round, and not conflating them, is the whole skill.

$$\text{number of words} = \underbrace{^n\mathrm{C}_r}_{\text{which letters}} \times \underbrace{r!}_{\text{in what order}}$$

The other recurring device is case-splitting: when a condition admits several possibilities (“at least $3$”, “either all or none”), enumerate the cases, count each, and add.

Key insight. “At least” and “at most” are handled in opposite ways. At least usually means listing the few cases above the threshold and adding them. At most is usually quicker as the complement — count everything, then subtract the cases that break the limit. Question 3 sets both against each other so that the difference is unmissable.

Question 1

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. DAUGHTER has $8$ distinct letters. Its vowels are A, U, E — three of them — and its consonants are D, G, H, T, R — five.

Stage 1 — select.

$$\text{vowels: } ^3\mathrm{C}_2 = 3, \qquad \text{consonants: } ^5\mathrm{C}_3 = 10$$

Stage 2 — arrange. The five chosen letters are all different, so they can be ordered in $5! = 120$ ways.

$$3 \times 10 \times 120 = 3600$$

Question 2

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. EQUATION has $8$ distinct letters: vowels E, U, A, I, O (five) and consonants Q, T, N (three).

Treat the vowels as one block and the consonants as another. There are two blocks, so they can be placed in $2!$ orders — vowels first or consonants first.

• Within the vowel block: $5! = 120$ arrangements.
• Within the consonant block: $3! = 6$ arrangements.

$$2! \times 5! \times 3! = 2 \times 120 \times 6 = 1440$$

The factor $2!$ is the one most often dropped — the question says the two groups each stay together, not which comes first.

Question 3

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) at least $3$ girls?   (iii) at most $3$ girls?

Solution. There are $13$ people in all and only $4$ girls, so the number of girls on the committee can be $0, 1, 2, 3$ or $4$.

(i) Exactly $3$ girls. Then $4$ boys are needed:

$$^4\mathrm{C}_3 \times {}^9\mathrm{C}_4 = 4 \times 126 = 504$$

(ii) At least $3$ girls. That means $3$ girls or $4$ girls — only two cases, since there are only four girls:

$$\underbrace{^4\mathrm{C}_3 \times {}^9\mathrm{C}_4}_{3 \text{ girls}} + \underbrace{^4\mathrm{C}_4 \times {}^9\mathrm{C}_3}_{4 \text{ girls}} = 504 + 1 \times 84 = 588$$

(iii) At most $3$ girls. This means $0$, $1$, $2$ or $3$ girls — four cases. It is quicker as a complement: everything except the single excluded case of $4$ girls.

$$^{13}\mathrm{C}_7 – {}^4\mathrm{C}_4 \times {}^9\mathrm{C}_3 = 1716 – 84 = 1632$$

Check: $588 + 1632 = 2220$, and the overlap (exactly $3$ girls) is $504$, so $2220 – 504 = 1716 = {}^{13}\mathrm{C}_7$ ✓

Question 4

If the different permutations of all the letters 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. EXAMINATION has $11$ letters: A ($2$), E, X, M, I ($2$), N ($2$), T, O. In alphabetical order the distinct letters are

$$\mathrm{A, E, I, M, N, O, T, X}$$

so A comes before E. Every word listed before the first E-word is therefore a word beginning with A, and there are no other letters in between.

Fix one A in the first position. The remaining $10$ letters are

$$\mathrm{A, E, X, M, I, N, A\!\!\!\!\phantom{x}, T, I, O, N} \to \mathrm{A}, \mathrm{E}, \mathrm{X}, \mathrm{M}, \mathrm{I}, \mathrm{I}, \mathrm{N}, \mathrm{N}, \mathrm{T}, \mathrm{O}$$

that is, ten letters in which I appears twice and N appears twice (only one A is left, so A is no longer repeated). Hence

$$\frac{10!}{2!\,2!} = \frac{3628800}{4} = 907200$$

Question 5

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. Divisible by $10$ means the number ends in $0$. That pins the last digit and, conveniently, removes the only digit that could not lead.

With $0$ fixed at the units place, the remaining five digits $1, 3, 5, 7, 9$ fill the other five positions in any order:

$$5! = 120$$

There is no leading-zero problem left to worry about, because the single $0$ has already been used.

Question 6

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. Select, then arrange.

$$^5\mathrm{C}_2 \times {}^{21}\mathrm{C}_2 \times 4! = 10 \times 210 \times 24 = 50400$$

Question 7

In an examination, a question paper consists of $12$ questions divided into two parts, 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. Let the student attempt $a$ questions from Part I and $8 – a$ from Part II. The constraints are

$$a \ge 3, \qquad 8 – a \ge 3, \qquad a \le 5, \qquad 8 – a \le 7$$

The first three give $3 \le a \le 5$, so there are exactly three cases.

The cases are mutually exclusive, so add:

$$210 + 175 + 35 = 420$$

Question 8

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. Split the deck into $4$ kings and $48$ non-kings.

$$^4\mathrm{C}_1 \times {}^{48}\mathrm{C}_4 = 4 \times 194580 = 778320$$

NCERT leaves the answer in the form $^4\mathrm{C}_1 \times {}^{48}\mathrm{C}_4$; the evaluated value is given above for completeness.

Question 9

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. Nine people sit in nine positions, numbered $1$ to $9$. The even positions are $2, 4, 6, 8$ — exactly four of them, which is precisely the number of women. So the assignment of groups to positions is forced:

• The $4$ women fill the $4$ even positions: $4! = 24$ ways.
• The $5$ men fill the $5$ odd positions $1, 3, 5, 7, 9$: $5! = 120$ ways.

$$5! \times 4! = 120 \times 24 = 2880$$

The count of even positions matching the count of women exactly is what makes the problem clean; there is no choosing to be done, only ordering.

Question 10

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. The condition splits the problem into two exclusive cases.

Case 1 — all three join. Their three places are taken, so $7$ more students are needed from the other $22$:

$$^{22}\mathrm{C}_7 = 170544$$

Case 2 — none of the three joins. All $10$ must come from the other $22$:

$$^{22}\mathrm{C}_{10} = 646646$$

The two cases cannot both happen, so add:

$$^{22}\mathrm{C}_7 + {}^{22}\mathrm{C}_{10} = 170544 + 646646 = 817190$$

NCERT prints the answer as the expression $^{22}\mathrm{C}_7 + {}^{22}\mathrm{C}_{10}$; the evaluated total is given above.

Question 11

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

Solution. ASSASSINATION has $13$ letters. Counting them:

$$\mathrm{A}\ (3), \quad \mathrm{S}\ (4), \quad \mathrm{I}\ (2), \quad \mathrm{N}\ (2), \quad \mathrm{T}\ (1), \quad \mathrm{O}\ (1)$$

$3 + 4 + 2 + 2 + 1 + 1 = 13$ ✓

Glue the four S’s into a single block. That block, together with the remaining nine letters, gives $10$ units:

$$\mathrm{A, A, A, I, I, N, N, T, O},\ [\mathrm{SSSS}]$$

Among these ten units, A appears three times, I twice and N twice:

$$\frac{10!}{3!\,2!\,2!} = \frac{3628800}{6 \times 2 \times 2} = \frac{3628800}{24} = 151200$$

The four S’s inside the block are identical, so no further factor is needed.

Common mistakes

• Question 1, arranging before selecting — or forgetting to arrange at all. The answer needs both $^3\mathrm{C}_2 \times {}^5\mathrm{C}_3$ and the $5!$. Stopping at $30$ counts letter-sets, not words.
• Question 2, omitting the $2!$. The vowel block and consonant block can appear in either order, so the two-block arrangement contributes a factor of $2$.
• Question 3(ii), including the case of $2$ girls. “At least $3$” means $3$ or more, and with only four girls available that is just two cases.
• Question 3(iii), enumerating all four cases. It works, but subtracting the one excluded case from $^{13}\mathrm{C}_7$ is far quicker and less error-prone.
• Question 4, forgetting that only one A remains. After fixing an A in the first position, the ten remaining letters contain a single A but still two I’s and two N’s. Dividing by $2!$ three times instead of twice halves the answer.
• Question 5, worrying about a leading zero. The divisibility condition puts $0$ at the end, so the leading position is automatically safe.
• Question 7, allowing $a = 2$ or $a = 6$. Part I has only $5$ questions and at least $3$ must come from each part, so $a$ is confined to $3$, $4$, $5$.
• Question 10, adding the two cases as a product. The student group either joins entirely or not at all — these are alternatives, so the counts add.
• Question 11, multiplying the S-block by $4!$. The S’s are identical; their internal order creates nothing new.

Practise next

• Exercise 7.1 — the next chapter, Binomial Theorem, where $^n\mathrm{C}_r$ becomes the coefficient in an expansion.
• Exercise 6.4 — worth working again alongside questions 3, 7 and 10, since all three are case-split combination problems.