# Combination

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

## Formula 1

### Formula

Combination nCr means
start from n, multiply r numbers,
and write r! in the denominator.

(n ≥ r)

Special values:
nCn = 1
nC1 = n
nC0 = 1

## Example 1

### Solution

7C3 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

## Formula 2

### Formula

nCr = nCn - r

Use this formula
when r seems to be greater than n/2.
This formula will save your time.

## Example 2

### Solution

6 seems to be greater than 8/2 = 4.

So change 8C6 to,
8 - 6 = 2,
8C2.

8C2 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

## Example 3

### Solution

nCr is used
when choosing r things from n things:
and no arranging.

The difference between
permutation nPr and combination nCr
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
5C2.

The number of ways
to choose 3 spoons from 6 spoons is
6C3.

The goal is to find the number of ways
to choose cups [and] spoons.

So [multiply] 6C3.

Rule of Product

So the number of ways
to choose 2 cups and 3 spoons is
5C26C3.

5C2 is,
start from 5, multiply 2 numbers,
5⋅4,
and write 2! in the denominator,
2⋅1.

6C3 is,
start from 6, multiply 3 numbers,
6⋅5⋅4,
and write 3! in the denominator,
3⋅2⋅1.

So 5C26C3
= [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