# Combination

How to solve a combination (nCr): 2 formulas, 3 examples, and their solutions.

## Formula_{n}C_{r}

Combination _{n}C_{r} means

start from n, multiply r numbers,

and write r! in the denominator.

(n ≥ r)

Special values:_{n}C_{n} = 1_{n}C_{1} = n_{n}C_{0} = 1

## Example_{7}C_{3}

_{7}C_{3} is,

start from 7, multiply 3 numbers,

7⋅6⋅5,

and write 3! in the denominator,

3⋅2⋅1.

Factorial

Cancel 6 in the numerator

and cancel the denominator 3⋅2⋅1.

7⋅5 = 35

So 35 is the answer.

## Formula_{n}C_{r} = _{n}C_{n - r}

_{n}C_{r} = _{n}C_{n - r}

Use this formula

when the given r is greater than n/2.

This formula will save your time.

## Example_{8}C_{6}

6 is greater than 8/2 = 4.

So change _{8}C_{6} to,

8 - 6 = 2,_{8}C_{2}.

_{8}C_{2} is,

start from 8, multiply 2 numbers,

8⋅7,

and write 2! in the denominator,

2⋅1.

Cancel the denominator 2⋅1

and reduce 8 in the numerator to, 8/2, 4.

4⋅7 = 28

So 28 is the answer.

## ExampleWord Problem

_{n}C_{r} is used

when choosing r things from n things:

and no arranging.

The difference between

permutation _{n}P_{r} and combination _{n}C_{r}

is the arranging part.

Permutation: Arranging it in a row

Combination: No arranging

So the number of ways

to choose 2 cups from 5 cups is_{5}C_{2}.

The number of ways

to choose 3 spoons from 6 spoons is_{6}C_{3}.

The goal is to find the number of ways

to choose cups [and] spoons.

So [multiply] _{6}C_{3}.

Rule of Product

So the number of ways

to choose 2 cups and 3 spoons is_{5}C_{2}⋅_{6}C_{3}.

_{5}C_{2} is,

start from 5, multiply 2 numbers,

5⋅4,

and write 2! in the denominator,

2⋅1.

_{6}C_{3} is,

start from 6, multiply 3 numbers,

6⋅5⋅4,

and write 3! in the denominator,

3⋅2⋅1.

So _{5}C_{2}⋅_{6}C_{3}

= [5⋅4]/[2⋅1] ⋅ [6⋅5⋅4]/[3⋅2⋅1].

Cancel the denominator 2⋅1

and reduce 4 in the numerator to, 4/2, 2.

Cancel the denominator 3⋅2⋅1

and cancel 6 in the numerator.

5⋅2 = 10

5⋅4 = 20

10⋅20 = 200

So 200 is the answer.