Permutations: Linear and Circular Arrangements | NIMCET

Quick answer: A permutation is an ordered arrangement. The number of ways to arrange r of n distinct objects is $^nP_r = \frac{n!}{(n - r)!}$ , and all n can be arranged in $n!$ ways. If some objects repeat (p alike, q alike, …), the count is $\frac{n!}{p!\,q!\cdots}$ . In a circle, n distinct objects can be arranged in $(n - 1)!$ ways, or $\frac{(n - 1)!}{2}$ if clockwise and anticlockwise are the same (as with a necklace).

Fundamental principles of counting

Principle	Rule
Multiplication	If one task can be done in m ways and then another in n ways, both can be done in $mn$ ways
Addition	If a task can be done in m ways or in n ways (not both), there are $m + n$ ways

Factorials

$n! = 1 \cdot 2 \cdot 3 \cdots n$ , with $0! = 1$ . Useful values: $5! = 120$ , $6! = 720$ , $7! = 5040$ .

Key formulas

Situation	Count
r of n distinct, no repetition	$^nP_r = \frac{n!}{(n - r)!}$
All n distinct	$n!$
r places, repetition allowed	$n^r$
n objects with p, q, r alike	$\frac{n!}{p!\,q!\,r!}$
Circular, n distinct	$(n - 1)!$
Circular, direction ignored (necklace)	$\frac{(n - 1)!}{2}$

Restricted arrangements

Restriction	Method
Certain objects always together	Treat them as one unit, then multiply by their internal arrangements
Certain objects never together	Total minus "always together"
No two of a group together	Arrange the others first, then place the group in the gaps
Fixed positions	Fill the restricted positions first

Worked examples

Example 1: nPr

In how many ways can 3 prizes be given to 8 students if no student gets more than one?

$^8P_3 = 8 \cdot 7 \cdot 6 = 336$ .

Example 2: identical letters

How many arrangements of the letters of BANANA are there?

6 letters with A three times and N twice: $\frac{6!}{3!\,2!} = 60$ .

Example 3: always together

In how many arrangements of the letters of MOTHER do the vowels come together?

Step 1: Treat O and E as one unit: 5 units, $5! = 120$ .

Step 2: The vowels can swap: $\times 2$ .

Answer: 240.

Example 4: no two together

In how many ways can 5 boys and 3 girls stand in a row so that no two girls are together?

Step 1: Arrange the boys: $5! = 120$ .

Step 2: There are 6 gaps; choose and arrange 3 girls in them: $^6P_3 = 120$ .

Answer: $120 \times 120 = 14400$ .

Example 5: circular

In how many ways can 6 people sit at a round table? $(6 - 1)! = 120$ .

Example 6: numbers from digits

How many 4-digit numbers can be formed from 0, 1, 2, 3, 4, 5 without repetition?

The first digit cannot be 0: $5 \times 5 \times 4 \times 3 = 300$ .

Example 7: rank of a word

Find the rank of CAT among the arrangements of its letters in dictionary order.

Step 1: Words starting with A: $2! = 2$ (ACT, ATC).

Step 2: Then CAT is the first word starting with C.

Answer: Rank 3.

Summary table

Idea	Key result
$^nP_r$	$\frac{n!}{(n - r)!}$
Repetition allowed	$n^r$
Identical objects	$\frac{n!}{p!\,q!\cdots}$
Circular	$(n - 1)!$
Necklace	$\frac{(n - 1)!}{2}$

Exam tips

Decide first whether order matters; if not, it is a combination problem.
Handle restrictions before filling unrestricted places.
"Not together" problems: use the gap method or subtract from the total.
For digit problems, remember 0 cannot be the first digit.

For practice, see NIMCET permutations and combinations PYQs and combinations and group division.

Common mistakes

Using $n!$ for a circle instead of $(n - 1)!$ .
Forgetting the internal arrangements of a grouped unit.
Dividing by 2 for circular seating of people, which applies only when direction does not matter.
Allowing 0 in the leading place of a number.

Frequently Asked Questions

What is the formula for nPr?

ⁿPᵣ = n!/(n − r)!.

How many ways can n people sit around a round table?

(n − 1)! ways.

How do you arrange letters with repeats?

Divide n! by the factorial of each repeat count. For BANANA, 6!/(3! 2!) = 60.

What is the difference between permutation and combination?

A permutation counts ordered arrangements; a combination counts selections where order does not matter.

What is 0!?

0! = 1.