Combinations (nCr)

Combinations (nCr)

How to solve combination problems (nCr): formulas, examples, and their solutions.

Formula: nPr

nCr is [starting from n and multiplying r factors] over [r!]

nCr is
[starting from n and multiplying r factors]
over [r!].

nPr = n! / [(n - r)! r!]

This formula can also be used
to solve nPr.

Example 1

Find the value of the given number. 7C3

7C3 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

There are 9 students in a class. Find the number of ways to choose 4 students.

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
9C4.

9C4 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

There are 5 boys and 6 girls in a class. Find the number of ways to choose 2 boys and 3 girls.

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
5C2 [ × ] 6C3.

5C2 is
starting from 5 and multiplying 2 factors
over 2!.

And 6C3 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: nCr = nCn - r

From n things, the number of ways to choose r things and the number of ways to remain (n - r) things are the same. So nCr = nC(n - r)

nCr 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 nC[r] = nC[n - r].

Example 4

Find the value of the given number. 8C6

8C6 = 8C8 - 6

So change 8C6 to 8C2.

It's obvious that
8C2 is easier to solve than 8C6.

This is the reason to change 8C6 to 8C2.

8C2 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: nC1

nC1 = n

nC1 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 nC1 = n.

Example 5

Find the value of the given number. 11C1

The latter number of 11C1 is 1.

So 11C1 = 11.

Formula: nC0

nC0 = 1

nC0 is defined as 1.

nC0 means
from n things,
[choose] 0 things with [no order].

There's only 1 way to do that:
not choosing anything.

So nC0 = 1.

Example 6

Find the value of the given number. 4C0

The latter number of 4C0 is 0.

So 4C0 = 1.