While you may be working with counting decisions and considering combinations and permutations, look up to their formulae and derivations could be worthy. As per mathematical considerations, permutation refers to the ordered or sequential arrangement of all the members in a set. However, combinations do not consider combination as a parameter. It is just one of the ways to select items from a collection or a set.

## Permutation

A permutation could be described as the choice of r objects from a set of n objects without replacement. Permutation considers maintaining the same order. The permutation formula is: $nP_{r}=\frac{n!}{(n-r)!}$

**Derivation **

The number of permutations for n different thing that may be taking r at a time is nPr. Now, let’s assume that there are r boxes which can hold one thing each. There could be as many permutations as the number of ways of filling in the vacant boxes (r) with objects (n).

No. of ways to fill the first box: n

No. of ways to fill the second box: (n – 1)

No. of ways to fill the third box: (n – 2)

No. of ways to fill the fourth box: (n – 3)

No. of ways to fill the rth box: (n – (r – 1))

Thus the no. of ways to fill r boxes in succession could be given as: n (n – 1) (n – 2) (n-3) . . . (n – (r – 1))

This could further be written as: n (n – 1) (n – 2) … (n – r + 1)

The no. permutations for n objects taken r at a time considering 0 < r £ n and objects aren’t repeated is n (n – 1) (n – 2) (n – 3) . . . (n – r + 1). ⇒ nPr= n ( n – 1) ( n – 2)( n – 3). . .( n – r + 1)

Further, multiplying the denominator and numerator by (n – r) (n – r – 1) . . . 3 × 2 × 1, we would get,

$nP_{r}=\frac{n!}{(n-r)!}$, where, 0 < r ≤ n.

## Combination

A combination could be described as the selection of r things from a set of n objects without replacement. Order is not considered in case of combinations.

**Derivation**

The number of combinations for n things, taken r at a time is denoted by nCr. Now, consider the following:

Ways to choose the first object from n objects: n

Ways to choose the second object from (n – 1) objects: (n – 1)

Ways of choosing the third object from (n – 2) objects: (n – 2)

Ways of choosing the fourth object from (n – 3) objects: (n – 3)

Ways to choose the rth object from (n-(r-1)) objects: (n-(r-1))

After the selection of r objects from the original set of n objects will result in an ordered subset including r elements. The no. of ways to select r items from a set of n elements would, therefore, be n (n – 1) (n – 2) (n-3) . . . (n – (r – 1)) or n (n – 1) (n – 2) … (n – r + 1)

Considering ordered subsets of r and all of its permutations, thus,

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

This is how the respective formulae gain their existence.