Factorial Notation and Permutations

The Factorial

Products like $4 \times 3 \times 2 \times 1$ appear so often in counting that they get their own symbol:

Key Point (Definition): $n! = 1 \times 2 \times 3 \times \cdots \times n$ for a natural number $n$, with the convention $0! = 1$. Immediately, $n! = n \times (n-1)!$.

The working skills:

• Cancel tails, never expand: $\frac{7!}{5!} = 7 \times 6 = 42$; $\frac{12!}{10!\,2!} = \frac{12 \times 11}{2} = 66$.
• Factor out the smaller factorial: $7! - 5! = 5!(7 \times 6 - 1) = 41 \times 5! = 4920$.
• Equations with factorials: from $\frac{1}{8!} + \frac{1}{9!} = \frac{x}{10!}$, multiply through by $10!$: $10 \times 9 + 10 = x$, so $x = 100$.

[Board Tip] $3! + 4! \ne 7!$ (it is $6 + 24 = 30$, while $7! = 5040$) — factorials do not add, multiply or distribute like ordinary powers. Exam setters plant this trap deliberately.

Permutations and the Formula for $^nP_r$

A permutation is an arrangement in a definite order of a number of objects taken some or all at a time. The 24 ROSE words are the permutations of 4 different letters taken all at a time.

Key Point (Theorems 1-2): The number of permutations of $n$ different objects taken $r$ at a time is

$$^nP_r = n(n-1)(n-2)\cdots(n-r+1) = \frac{n!}{(n-r)!}, \quad 0 \le r \le n$$

when repetition is not allowed — and $n^r$ when repetition is allowed.

The derivation is the vacant-places method run in general: the first place has $n$ choices, the next $n - 1$, down to $n - r + 1$ for the $r$-th place; multiplying and rewriting the product as a quotient of factorials gives the formula. Special values: $^nP_n = n!$, $^nP_1 = n$, $^nP_0 = 1$.

Quick illustrations: 3-letter words from NUMBER $= {^6P_3} = 120$ (with repetition: $6^3 = 216$); a Chairman and Vice-Chairman from 12 people $= {^{12}P_2} = 132$ — order matters because the posts differ.

[JEE Tip] Equations in $^nP_r$ reduce to polynomial equations after cancelling common factors — e.g. $^nP_5 = 42\,{^nP_3}$ gives $(n-3)(n-4) = 42$, so $n = 10$. Always discard roots that violate $n \ge r$ or negativity — the algebra offers them, the counting rejects them.

Solved Examples

Example 1: Evaluate

Evaluate (i) $5!$ (ii) $7!$ (iii) $7! - 5!$.

Solution:

Step 1 — (i) and (ii) by the definition. $5! = 120$ and $7! = 5040$.

Step 2 — (iii) Factor out the smaller factorial. $7! - 5! = 5!(7 \times 6 - 1) = 5! \times 41$.

Step 3 — Evaluate. $120 \times 41 = 4920$ (or directly $5040 - 120$).

Takeaway: Factor the smaller factorial out of sums and differences — never expand both.

Example 2: Compute quotients

Compute (i) $\frac{7!}{5!}$ (ii) $\frac{12!}{10!\,2!}$.

Solution:

Step 1 — (i) Cancel the common tail. $\frac{7!}{5!} = 7 \times 6 = 42$.

Step 2 — (ii) Cancel, then divide. $\frac{12!}{10!\,2!} = \frac{12 \times 11}{2} = 66$.

Takeaway: Quotients of factorials are falling products — expansion is never needed.

Example 3: The shape of $^nC_r$

Evaluate $\frac{n!}{r!\,(n-r)!}$ for $n = 5$, $r = 2$.

Solution:

Step 1 — Substitute. $\frac{5!}{2!\,3!}$.

Step 2 — Cancel the 3! tail. $\frac{5 \times 4}{2!} = \frac{20}{2} = 10$.

Takeaway: This expression is the combination count $^5C_2$ — the formula arrives before its name.

Example 4: Factorial equation

If $\frac{1}{8!} + \frac{1}{9!} = \frac{x}{10!}$, find $x$.

Solution:

Step 1 — Multiply through by $10!$. $\frac{10!}{8!} + \frac{10!}{9!} = x$.

Step 2 — Cancel each quotient. $10 \times 9 + 10 = 90 + 10$.

Step 3 — Conclude. $x = 100$.

Takeaway: Clearing by the LARGEST factorial turns the equation into simple arithmetic.

Example 5: Same trick again

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

Solution:

Step 1 — Multiply by $8!$. $\frac{8!}{6!} + \frac{8!}{7!} = x$.

Step 2 — Evaluate. $8 \times 7 + 8 = 56 + 8 = 64$, so $x = 64$.

Takeaway: The same clearing move works for any such pair of consecutive reciprocals.

Example 6: Evaluate $\frac{n!}{(n-r)!}$

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

Solution:

Step 1 — (i) Two falling factors. $\frac{6!}{4!} = 6 \times 5 = 30$.

Step 2 — (ii) Five falling factors. $\frac{9!}{4!} = 9 \times 8 \times 7 \times 6 \times 5 = 15120$.

Takeaway: $\frac{n!}{(n-r)!}$ is a product of exactly r falling factors starting at n — the permutation formula in the making.

Example 7: NUMBER words

How many 3-letter words (with or without meaning) can be formed from the letters of NUMBER (i) without repetition (ii) with repetition?

Solution:

Step 1 — (i) Apply the permutation formula. NUMBER has 6 distinct letters: $^6P_3 = 6 \times 5 \times 4 = 120$.

Step 2 — (ii) With repetition every place stays full. $6^3 = 216$.

Takeaway: Same six letters, two settings — the repetition clause switches between $^nP_r$ and $n^r$.

Example 8: Two posts

In how many ways can a Chairman and a Vice-Chairman be chosen from 12 persons, no person holding both posts?

Solution:

Step 1 — Recognise that order matters. The posts are DIFFERENT — (A chair, B vice) is not (B chair, A vice).

Step 2 — Apply the formula. $^{12}P_2 = 12 \times 11 = 132$.

Step 3 — Contrast. Choosing an unordered pair of equal committee members would be $\frac{132}{2} = 66$ — the permutation-combination divide previewed.

Takeaway: Distinct posts = ordered arrangement; identical roles = unordered selection.

Example 9: Solve for $n$

Find $n$ such that (i) $^nP_5 = 42\,{^nP_3}$, $n > 4$ (ii) $\frac{^nP_4}{^{n-1}P_4} = \frac{5}{3}$, $n > 4$.

Solution:

Step 1 — (i) Write both sides as falling products. $n(n-1)(n-2)(n-3)(n-4) = 42\,n(n-1)(n-2)$.

Step 2 — Cancel the common non-zero factors. $(n-3)(n-4) = 42$.

Step 3 — Solve the quadratic. $n^2 - 7n - 30 = 0$ gives $(n - 10)(n + 3) = 0$: $n = 10$ (reject $-3$ — counts cannot be negative).

Step 4 — (ii) Same strategy. $3n(n-1)(n-2)(n-3) = 5(n-1)(n-2)(n-3)(n-4)$; cancelling gives $3n = 5(n-4)$, so $2n = 20$, $n = 10$.

Takeaway: Cancel the shared falling factors first; the leftover equation is tiny. Discard roots the counting context forbids.

Example 10: Solve for $r$

Find $r$ if $5\,{^4P_r} = 6\,{^5P_{r-1}}$.

Solution:

Step 1 — Write in factorial form. $5 \cdot \frac{4!}{(4-r)!} = 6 \cdot \frac{5!}{(6-r)!}$.

Step 2 — Expand the larger denominator. $(6-r)! = (6-r)(5-r)(4-r)!$, so the equation collapses to $(6-r)(5-r) = 6$.

Step 3 — Solve. $r^2 - 11r + 24 = 0$ gives $r = 3$ or $r = 8$.

Step 4 — Domain check. The symbols need $r \le 4$ and $r - 1 \le 5$: only $r = 3$ survives.

Takeaway: The algebra offers extra roots; the symbol domains ($0 \le r \le n$) reject them.

Example 11: A ratio

Find $n$ if $^{n-1}P_3 : {^nP_4} = 1 : 9$.

Solution:

Step 1 — Relate the two symbols. $^nP_4 = n(n-1)(n-2)(n-3) = n \cdot {^{n-1}P_3}$.

Step 2 — Form the ratio. $\frac{^{n-1}P_3}{^nP_4} = \frac{1}{n} = \frac{1}{9}$.

Step 3 — Conclude. $n = 9$.

Takeaway: Spotting the one-factor relationship $^nP_4 = n \cdot {^{n-1}P_3}$ makes this a one-liner.

Example 12: Two little equations

Find $r$ if (i) $^5P_r = 2\,{^6P_{r-1}}$ (ii) $^5P_r = {^6P_{r-1}}$.

Solution:

Step 1 — (i) Factorial form and collapse. $\frac{5!}{(5-r)!} = \frac{2 \cdot 6!}{(7-r)!}$; with $(7-r)! = (7-r)(6-r)(5-r)!$: $(7-r)(6-r) = 12$.

Step 2 — Solve and check. $r^2 - 13r + 30 = 0$ gives $r = 10$ or $r = 3$; domain $r \le 5$ keeps $r = 3$.

Step 3 — (ii) Same route. $(7-r)(6-r) = 6$ gives $r^2 - 13r + 36 = 0$: $r = 9$ or $r = 4$; domain keeps $r = 4$.

Takeaway: Both parts ride the same expansion $(7-r)! = (7-r)(6-r)(5-r)!$ — set it up once, reuse it.