# Combinations (_{n}C_{r})

How to solve combination problems (_{n}C_{r}): formulas, examples, and their solutions.

## Formula: _{n}P_{r}

_{n}C_{r} is

[starting from *n* and multiplying *r* factors]

over [*r*!].

_{n}P_{r} = *n*! / [(*n* - *r*)! *r*!]

This formula can also be used

to solve _{n}P_{r}.

## Example 1

_{7}C_{3} is

starting from 7 and multiplying 3 factors

over 3!.

3! = 3⋅2⋅1

3⋅2⋅1 = 6

So cancel 6 in the numerator

and cancel 3⋅2⋅1 in the denominator.

Then the right side is 7⋅5.

7⋅5 = 35

So 35 is the answer.

## Example 2

It says

from 9 students,

choose 4 students.

And there's [no order].

When there's only [choosing] and [no order],

then use the combination.

So the number of ways to choose is_{9}C_{4}.

_{9}C_{4} is

starting from 9 and multiplying 4 factors

over 4!.

4! = 4⋅3⋅2⋅1

Cancel 4 in the denominator

and reduce 8 in the numerator to 2.

Cancel 6 in the numerator

and cancel 3⋅2⋅1.

Then the right side is 9⋅2⋅7.

9⋅2 = 18

18⋅7 = 126

So 126 is the answer.

## Example 3

It says

from 5 boys,

choose 2 boys.

There's [no order].

And it says

from 6 girls,

choose 3 girls.

Also there's [no order].

When there's only [choosing] and [no order],

then use the combination.

So the number of ways to choose

2 boys [and] 3 girls is_{5}C_{2} [ × ] _{6}C_{3}.

_{5}C_{2} is

starting from 5 and multiplying 2 factors

over 2!.

And _{6}C_{3} is

starting from 6 and multiplying 3 factors

over 3!.

2! = 2⋅1

3! = 3⋅2⋅1

Cancel 2⋅1 in the denominator

and reduce 4 in the numerator to 2.

Cancel 6 in the numerator

and cancel 3⋅2⋅1.

Then the right side is 5⋅2 ⋅ 5⋅4.

5⋅2 = 10

5⋅4 = 20

10⋅20 = 200

So 200 is the answer.

## Formula: _{n}C_{r} = _{n}C_{n - r}

_{n}C_{r} means

the number of ways

to [choose *r*] things with no order

from *n*.

This also means

the number of ways

to [remain (*n* - *r*)] things with no order.

So the number of ways to [choose *r*] things

and the number of ways to [remain (*n* - *r*)] things

are the same.

So _{n}C_{[r]} = _{n}C_{[n - r]}.

## Example 4

_{8}C_{6} = _{8}C_{8 - 6}

So change _{8}C_{6} to _{8}C_{2}.

It's obvious that_{8}C_{2} is easier to solve than _{8}C_{6}.

This is the reason to change _{8}C_{6} to _{8}C_{2}.

_{8}C_{2} is

starting from 8 and multiplying 2 factors

over 2!.

2! = 2⋅1

Cancel 2 in the denominator

and reduce 8 in the numerator to 4.

Then the right side is 4⋅7.

4⋅7 = 28

So 28 is the answer.

## Formula: _{n}C_{1}

_{n}C_{1} is

starting from *n* and multiplying 1 factor

over 1!.

This is *n*/1!.

Recall that 1! = 1.

Factorial - Example 3

So *n*/1! = *n*.

So _{n}C_{1} = *n*.

## Example 5

The latter number of _{11}C_{1} is 1.

So _{11}C_{1} = 11.

## Formula: _{n}C_{0}

_{n}C_{0} is defined as 1._{n}C_{0} means

from *n* things,

[choose] 0 things with [no order].

There's only 1 way to do that:

not choosing anything.

So _{n}C_{0} = 1.

## Example 6

The latter number of _{4}C_{0} is 0.

So _{4}C_{0} = 1.