NCERT Class 11 Mathematics — Permutations and Combinations, Exercise 6.2. All 5 questions solved.

Factorial notation compresses the falling products of the previous exercise:

$$n! = n \times (n-1) \times (n-2) \times \cdots \times 3 \times 2 \times 1, \qquad 0! = 1$$

The property that makes factorials workable is that each contains the next:

$$n! = n \times (n-1)!$$

so $8! = 8 \times 7 \times 6!$ and a quotient like $\dfrac{8!}{6!}$ collapses to $8 \times 7$ without ever computing either factorial.

Key insight. Never evaluate a factorial you are about to cancel. In question 3, $\dfrac{8!}{6!\,2!}$ becomes $\dfrac{8 \times 7}{2}$ in one line; working out $40320$ and $720$ first is slower and invites arithmetic slips. The rule is to expand the larger factorial only as far down as the smaller one, then stop.

Question 1

Evaluate (i) $8!$    (ii) $4! – 3!$

Solution.

(i)

$$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$$

(ii) $4! = 24$ and $3! = 6$, so

$$4! – 3! = 24 – 6 = 18$$

Note that $4! – 3!$ is not $(4-3)! = 1!$. Factorials do not distribute over subtraction, which is exactly the point question 2 makes about addition.

Question 2

Is $3! + 4! = 7!$?

Solution. Compute the two sides separately.

$$3! + 4! = 6 + 24 = 30$$

$$7! = 5040$$

Since $30 \ne 5040$, the statement is false.

The temptation is to read the factorial sign as though it distributed over addition. It does not: $a! + b! \ne (a+b)!$, and here the two sides differ by a factor of $168$.

Question 3

Compute $\dfrac{8!}{6! \times 2!}$

Solution. Write $8!$ down only as far as $6!$:

$$8! = 8 \times 7 \times 6!$$

so the $6!$ cancels:

$$\frac{8!}{6! \times 2!} = \frac{8 \times 7 \times 6!}{6! \times 2!} = \frac{8 \times 7}{2 \times 1} = \frac{56}{2} = 28$$

Question 4

If $\dfrac{1}{6!} + \dfrac{1}{7!} = \dfrac{x}{8!}$, find $x$.

Solution. Multiply every term by $8!$ — this clears all three denominators at once and leaves $x$ alone on the right.

$$\frac{8!}{6!} + \frac{8!}{7!} = x$$

Now expand each quotient only as far as needed:

$$\frac{8!}{6!} = 8 \times 7 = 56, \qquad \frac{8!}{7!} = 8$$

$$x = 56 + 8 = 64$$

Question 5

Evaluate $\dfrac{n!}{(n-r)!}$, when (i) $n = 6$, $r = 2$    (ii) $n = 9$, $r = 5$

Solution. The quotient $\dfrac{n!}{(n-r)!}$ is the product of the top $r$ factors of $n!$ — it is $^n\mathrm{P}_r$, the count from Exercise 6.1 written in factorial form.

(i) $n – r = 4$, so

$$\frac{6!}{4!} = 6 \times 5 = 30$$

(ii) $n – r = 4$, so

$$\frac{9!}{4!} = 9 \times 8 \times 7 \times 6 \times 5 = 15120$$

Five factors, because $r = 5$ — the count of surviving factors is always $r$, which is a useful check that you have stopped in the right place.

Common mistakes

• Question 1(ii), computing $(4-3)!$. Factorial applies to each term separately: $4! – 3! = 24 – 6$, not $1! = 1$.
• Question 2, answering “yes” from the pattern $3 + 4 = 7$. The bases add but the factorials do not. This question exists solely to break that intuition.
• Question 3, evaluating all three factorials. $8! = 40320$, $6! = 720$ and $2! = 2$ give the same answer, but through numbers large enough to mis-key. Cancelling first is both faster and safer.
• Question 4, adding the fractions over a common denominator of $6! \times 7!$. That denominator is enormous and unnecessary. Multiplying through by $8!$ — the denominator already on the right — is the efficient move.
• Question 5, expanding too far or not far enough. $\dfrac{9!}{4!}$ keeps five factors ($9$ down to $5$), not four. Counting them against $r$ catches the error immediately.
• Writing $0! = 0$. By convention $0! = 1$, which is what makes $^n\mathrm{P}_n = \dfrac{n!}{0!} = n!$ come out correctly.

Practise next

• Exercise 6.3 — permutations, where $\dfrac{n!}{(n-r)!}$ of question 5 is named $^n\mathrm{P}_r$ and applied to words and numbers.
• Exercise 6.4 — combinations, where the same factorials appear with an extra $r!$ in the denominator.