In mathematics, a combination is a way of selecting items from a collection where the order of selection does not matter. In smaller cases, it is possible to count the number of combinations.

**What is a combination? **

Combination is defined as “An arrangement of objects where the order in which the objects are selected does not matter.” Combination means “Selection of things”, where the order of things has no importance.

For example, if we want to buy a milkshake and we are allowed to combine any 3 flavors from Apple, Banana, Cherry, and Durian, then the combination of Apple, Banana, and Cherry is the same as the combination Banana, Apple, Cherry.

So if we are supposed to make a combination out of these possible flavors, then firstly, let us shorten the name of the fruits by selecting the first letter of their names. We only have 4 possible combinations for the question above ABC, ABD, ACD, and BCD. Also, do notice that these are the only possible combination. This can be easily understood by the combination Formula.

The Combination of 4 objects taken 3 at a time are the same as the number of subgroups of 3 objects taken from 4 objects.

Take another example, given three fruits; say an apple, an orange, and a pear, three combinations of two can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange. More formally, a *k*-combination of a set is a subset of *k* distinct elements of *S*. If the set has *n* elements, the number of *k*-combinations is equal to the binomial coefficient

**\(^{n}C_{k}= \frac{(n)(n-1)(n-2)……(n-k+1)}{(k-1)(k-2)…….(1)}\)**

** **

which can be written as

**\(\frac{n!}{k!(n-k)!}\), when n>k**

**and 0, when n<k**

Where, n = distinct object to choose from

C = Combination

K = spaces to fill (Where k can be replaced by r also)

Combination Formula can be written as:

**\(^{n}C_{r}, _nC_{r}, C(n,r), C_{r}^{n}\)**

**Theorems- **

**Relation between Permutation and Combination-**

**Theorem :** \(^{n}P_{r} = ^{n}C_{r}. r!\)

Corresponding to each combination of \(^{n}C_{r}\), we have r! permutations because r objects in every combination can be rearranged in r! ways.

**Proof-**

\(^{n}P_{r} = ^{n}C_{r}.r!\)

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

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

Hence the theorem states true.

**Theorem :** \(^{n}C_{r}+ ^{n}C_{r-1}= ^{n+1}C_{r}\)

Proof –

\(^{n}C_{r}+ ^{n}C_{r-1}= \frac{n!}{r!(n-r)!}+ \frac{n!}{(r-1)! (n-r+1)!}\)

\(= \frac{n!}{r(r-1)!(n-r)!}+ \frac{n!}{(r-1)! (n-r+1)(n-r)!}\)

\(= \frac{n!}{(r-1)!(n-r)!} \left [ \frac{1}{r} + \frac{1}{(n-r+1)} \right ]\)

\(= \frac{n!}{(r-1)!(n-r)!} \left [ \frac{n + 1}{r (n-r+1)} \right ]\)

\(= \frac{(n+1)!}{(r)!(n+1-r)!}\)

\(= ^{n+1}C_{r}\)

Let’s Work Out-
Thus ST= TS, TU = UT, and SU=US. Thus we have 3 ways of team selection. By combination formula we have- \(^{3}C_{2} = \frac{3!}{2! (3-2)!}\) \(= \frac{3 \times 2 \times 1}{2 \times 1 \times 1} =3\) |

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