Feedback form

12 Choose 2

Combination Calculator

What is 12 choose 2?

Answer

12 choose 2 is

66

Order doesn't matter, no repeats: 66 ways to choose 2 of 12 items.

C(12, 2) = 12! / (2! · 10!)

= 66

If order matters (permutations): P(12, 2) = 12 × 11 = 132 — each combination in 2! = 2 orders.

Row 12 of Pascal's triangle
k	12 choose k
0	1
1	12
2	66
3	220
4	495
5	792
6	924
7	792
8	495
9	220
10	66
11	12
12	1

Row sum: 2¹² = 4,096 — all subsets of 12 items.

Found an error in the calculation or the information?

See Also

Random Number Generator - Pick a random whole number between two bounds
Percent to Probability Converter - Convert Percents to "1 in X" Probability Chance
Factors of a Number - List all Factors and Factor Pairs of a Number

Share This Calculation

Print

https://calculat.io/en/number/combinations/12--2

<a href="https://calculat.io/en/number/combinations/12--2">12 Choose 2: Combinations C(12, 2) - Calculatio</a>

Facebook

Twitter

Telegram

WhatsApp

Viber

Email

12 Choose 2: Combinations C(12, 2). 12 choose 2 = 66: the number of ways to choose 2 of 12 items when order doesn't matter and nothing repeats. C(12, 2) = (12 × 11) / (2 × 1).

https://calculat.io/en/number/combinations/12--2/generated.jpg

n choose 2 for other n

n choose 2	Answer
5 choose 2	10
6 choose 2	15
7 choose 2	21
8 choose 2	28
9 choose 2	36
10 choose 2	45
11 choose 2	55
12 choose 2	66
13 choose 2	78
14 choose 2	91
15 choose 2	105
16 choose 2	120
17 choose 2	136
18 choose 2	153
19 choose 2	171

Recalculate with different values ↑

FAQ

What is 12 choose 2?

12 choose 2 is 66

Why is 12 choose 2 the same as 12 choose 10?

Every way to choose 2 of 12 items leaves exactly 10 behind, so choosing 2 and choosing which 10 to leave out are the same count: 12 choose 2 = 12 choose 10 = 66. In the formula n! / (k! (n − k)!) the two factorials below just swap places.

How many ways are there if order matters?

P(12, 2) = 132. Each of the 66 combinations can be put in order in 2! = 2 ways, so permutations = combinations × k!: P(n, k) = n! / (n − k)!.