In mathematics, permutation refers to the arrangement of all the members of a set in some order or sequence, while combination does not regard order as a parameter. It is just a way of selecting items from a set or collection.

**Permutation Formula: **A permutation is the choice of *r* things from a set of *n* things without replacement. Order matters in permutation.

\(nP_{r}=\frac{n!}{(n-r)!}\)

**Combination Formula: **A combination is the choice of *r* things from a set of *n* things without replacement. Order does not matter in combination.

\(nC_{r}=\frac{n!}{(n-r)!r!}=\frac{nPr}{r!}\)

**Derivation:**

**Number of permutations of n different things taking r at a time is**nPr

Let us assume that there are r boxes and each of them can hold one thing. There will be as many permutations as there are ways of filling in *r *vacant boxes by *n* objects.

No. of ways first box can be filled: *n*

No. of ways second box can be filled: (*n *– 1)

No. of ways third box can be filled: (*n *– 2)

No. of ways fourth box can be filled: (*n *– 3)

No. of ways *r*th box can be filled: (*n *– (*r *– 1))

Therefore, no. of ways of filling in *r *boxes in succession can be given by:

*n *(*n *– 1) (*n *– 2) (n-3) . . . (*n *– (*r *– 1))

This can be written as:

*n *(*n *– 1) (*n *– 2) … (*n *– *r *+ 1)

The no. of permutations of *n *different objects taken *r *at a time, where 0 < *r *£ *n *and the objects do not repeat is *n *(n – 1) (n – 2) (n – 3) . . . (n – *r *+ 1).

⇒ nPr= n ( n – 1) ( n – 2)( n – 3). . .( n – r + 1)

Multiplying numerator and denominator by (*n *– *r*) (*n *– *r *– 1) . . . 3 × 2 × 1, we get

\(nP_{r}=\frac{[n(n-1)(n-2)(n-3)…(n-r+1)(n-r)(n-r-1)..3\times 2\times 1]}{(n-r)(n-r-1)..3\times 2\times 1}=\frac{n!}{(n-r)!}\)

Hence,

\(nP_{r}=\frac{n!}{(n-r)!}\)

Where 0 < r ≤n

**Number of combinations of n distinct things taking r at a time is**nCr

No. of ways to select first object from *n* distinct objects: *n* ways

No. of ways to select second object from (*n-1)* distinct objects: (*n-1)* ways

No. of ways to select third object from (*n-2)* distinct objects: (*n-2)* ways

No. of ways to select fourth object from (*n-3)* distinct objects: (*n-3)* ways

No. of ways to select rth object from (*n-(r-1))* distinct objects: (*n-(r-1))* ways

Completing selection *r* things of the original set of n things creates an ordered sub-set of *r* elements.

∴ the number of ways to make a selection of r elements of the original set of *n *elements is *n *(*n *– 1) (*n *– 2) (n-3) . . . (*n *– (*r *– 1)) or *n *(*n *– 1) (*n *– 2) … (*n *– *r *+ 1)

Let us consider the ordered sub-set of *r* elements and all its permutations. The total number of all permutations of this sub-set is equal to r! because *r *objects in every combination can be rearranged in *r! *ways.

Hence, the total number of permutations of *n *different things taken *r *at a time is nCr ×r! On the other hand, it is nPr**.**

**\(nP_{r}=nC_{r}\times r!\)**

\(nC_{r}=\frac{nP_{r}}{r!}=\frac{n!}{(n-r)!r!}\)

**Permutations and Combinations in Real Life**

Permutations and combinations are techniques which help us to answer the question or determine the number of different ways of arranging and selecting objects without actually listing them in real life. For example, when you have to arrange people, pick a team captain, pick two favorite colors, in order, from a color brochure, or selection of menu, food, clothes, subjects, team, etc.

**Solved Examples**

**Questions 1:** Evaluate

- 12P2
- 10C3

**Solution:**

- Here, n= 12 and r= 2

\(12P_{2}=\frac{12!}{(12-2)!}=132\)

- Here, n= 10 and r = 3

\(10C_{3}=\frac{10!}{(10-3)!3!}=\frac{10!}{7!3!}=120\)

**Questions 2:** Teacher asks a student to choose 6 items from the table. If the table has 20 items to choose, how many ways could the students choose the things?

**Solution:** Here, student has to choose 6 items from 20 items. Here, r= 6 and n= 20

Combination,

\(nC_{r}=\frac{n!}{(n-r)!r!}\)

\(20C_{6}=\frac{20!}{(20-6)!6!}=38760\)

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