NCERT Class 11 Mathematics — Permutations and Combinations, Exercise 6.4. All 9 questions solved.

A combination is a selection, so order does not matter. Choosing $r$ objects from $n$ distinct objects can be done in

$$^n\mathrm{C}_r = \frac{n!}{r!\,(n-r)!} = \frac{^n\mathrm{P}_r}{r!}$$

ways — the permutation count divided by $r!$, because each selection was counted once for every way of ordering it.

Two properties are used constantly:

$$^n\mathrm{C}_r = {}^n\mathrm{C}_{n-r}, \qquad ^n\mathrm{C}_a = {}^n\mathrm{C}_b \Longrightarrow a = b \ \text{ or } \ a + b = n$$

The second is the whole of question 1, and its “or” is what makes the question non-trivial.

Key insight. When a selection has to satisfy several conditions at once — so many of this, so many of that — choose each group separately and multiply. Questions 4, 5, 7 and 8 are all that pattern. When the conditions are alternatives instead (“either this or that”), add rather than multiply.

Question 1

If $^n\mathrm{C}_8 = {}^n\mathrm{C}_2$, find $^n\mathrm{C}_2$.

Solution. Two combinations with the same $n$ are equal in exactly two circumstances: the lower indices are equal, or they sum to $n$. Here $8 \ne 2$, so

$$8 + 2 = n \quad\Longrightarrow\quad n = 10$$

Therefore

$$^{10}\mathrm{C}_2 = \frac{10 \times 9}{2 \times 1} = 45$$

Question 2

Determine $n$ if (i) $^{2n}\mathrm{C}_3 : {}^n\mathrm{C}_3 = 12 : 1$    (ii) $^{2n}\mathrm{C}_3 : {}^n\mathrm{C}_3 = 11 : 1$

Solution. Expand both combinations as products, which is far cleaner than factorials here:

$$^{2n}\mathrm{C}_3 = \frac{2n(2n-1)(2n-2)}{6}, \qquad ^n\mathrm{C}_3 = \frac{n(n-1)(n-2)}{6}$$

Taking the ratio, the $6$’s cancel and $2n(2n-2) = 4n(n-1)$:

$$\frac{^{2n}\mathrm{C}_3}{^n\mathrm{C}_3} = \frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = \frac{4n(n-1)(2n-1)}{n(n-1)(n-2)} = \frac{4(2n-1)}{n-2}$$

(i) Set this equal to $12$:

$$4(2n-1) = 12(n-2) \quad\Longrightarrow\quad 8n – 4 = 12n – 24 \quad\Longrightarrow\quad 20 = 4n \quad\Longrightarrow\quad n = 5$$

(ii) Set it equal to $11$:

$$4(2n-1) = 11(n-2) \quad\Longrightarrow\quad 8n – 4 = 11n – 22 \quad\Longrightarrow\quad 18 = 3n \quad\Longrightarrow\quad n = 6$$

Question 3

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

Solution. A chord is determined by its two endpoints, and the chord from $A$ to $B$ is the same as the chord from $B$ to $A$ — so this is a selection, not an arrangement:

$$^{21}\mathrm{C}_2 = \frac{21 \times 20}{2} = 210$$

(No three points on a circle are collinear, so every pair really does give a distinct chord.)

Question 4

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

Solution. The two selections are independent, so choose each and multiply:

$$^5\mathrm{C}_3 \times {}^4\mathrm{C}_3 = 10 \times 4 = 40$$

Question 5

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. Three independent selections, one per colour:

$$^6\mathrm{C}_3 \times {}^5\mathrm{C}_3 \times {}^5\mathrm{C}_3 = 20 \times 10 \times 10 = 2000$$

Question 6

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

Solution. Split the deck into the $4$ aces and the $48$ non-aces, and choose from each group separately.

• One ace from four: $^4\mathrm{C}_1 = 4$
• Four non-aces from forty-eight: $^{48}\mathrm{C}_4 = \dfrac{48 \times 47 \times 46 \times 45}{24} = 194580$

$$4 \times 194580 = 778320$$

The word exactly is what forces the second group to be drawn from the non-aces: choosing four more cards from the whole remaining $51$ would allow a second ace.

Question 7

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. The $17$ players split into $5$ bowlers and $12$ non-bowlers. The team needs $4$ bowlers and therefore $11 – 4 = 7$ non-bowlers:

$$^5\mathrm{C}_4 \times {}^{12}\mathrm{C}_7 = 5 \times 792 = 3960$$

($^{12}\mathrm{C}_7 = {}^{12}\mathrm{C}_5 = 792$, which is the quicker way to compute it.)

Question 8

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.

$$^5\mathrm{C}_2 \times {}^6\mathrm{C}_3 = 10 \times 20 = 200$$

Question 9

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. The two compulsory courses are not chosen — they are given. So they occupy two of the five places automatically, and the real question is how to fill the remaining

$$5 – 2 = 3 \text{ places}$$

from the

$$9 – 2 = 7 \text{ remaining courses}$$

$$^7\mathrm{C}_3 = \frac{7 \times 6 \times 5}{6} = 35$$

Common mistakes

• Question 1, concluding $n = 8$ or $n = 2$. $^n\mathrm{C}_8 = {}^n\mathrm{C}_2$ holds because $8 + 2 = n$, giving $n = 10$. The identity $^n\mathrm{C}_r = {}^n\mathrm{C}_{n-r}$ is the reason, and forgetting it makes the question look unsolvable.
• Question 3, treating a chord as ordered. $^{21}\mathrm{P}_2 = 420$ counts each chord twice, once from each end.
• Questions 4, 5, 7 and 8, adding instead of multiplying. The groups are chosen simultaneously, so the counts multiply. Addition is for alternatives, not for simultaneous requirements.
• Question 6, choosing the other four cards from $51$. That permits a second ace and breaks the word “exactly”. The remaining four must come from the $48$ non-aces.
• Question 7, forgetting how many non-bowlers are needed. Eleven players minus four bowlers is seven, not six. Writing the two numbers down before computing prevents the slip.
• Question 9, choosing $5$ from $9$. The two compulsory courses are already in, so only $3$ places remain and only $7$ courses are eligible for them. Answering $^9\mathrm{C}_5 = 126$ ignores the word “compulsory” entirely.
• Confusing $^n\mathrm{P}_r$ with $^n\mathrm{C}_r$. The test is whether re-ordering the chosen objects produces a genuinely different outcome. For a team or a hand of cards it does not; for a chairman-and-vice-chairman pair it does.

Practise next

• Miscellaneous Exercise on Chapter 6 — problems mixing selection with arrangement, where a combination is followed by a permutation of what was chosen.
• Exercise 7.1 — the next chapter, Binomial Theorem, in which $^n\mathrm{C}_r$ reappears as the binomial coefficient.