Feedback form

15 Choose 2

Combination Calculator

What is 15 choose 2?

Answer

15 choose 2 is

105

Order doesn't matter, no repeats: 105 ways to choose 2 of 15 items.

C(15, 2) = 15! / (2! · 13!)

= 105

If order matters (permutations): P(15, 2) = 15 × 14 = 210 — each combination in 2! = 2 orders.

Row 15 of Pascal's triangle
k	15 choose k
0	1
1	15
2	105
3	455
4	1,365
5	3,003
6	5,005
7	6,435
8	6,435
9	5,005
10	3,003
11	1,365
12	455
13	105
14	15
15	1

Row sum: 2¹⁵ = 32,768 — all subsets of 15 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/15--2

<a href="https://calculat.io/en/number/combinations/15--2">15 Choose 2: Combinations C(15, 2) - Calculatio</a>

Facebook

Twitter

Telegram

WhatsApp

Viber

Email

15 Choose 2: Combinations C(15, 2). 15 choose 2 = 105: the number of ways to choose 2 of 15 items when order doesn't matter and nothing repeats. C(15, 2) = (15 × 14) / (2 × 1).

https://calculat.io/en/number/combinations/15--2/generated.jpg

n choose 2 for other n

n choose 2	Answer
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
20 choose 2	190

Recalculate with different values ↑

FAQ

What is 15 choose 2?

15 choose 2 is 105

Why is 15 choose 2 the same as 15 choose 13?

Every way to choose 2 of 15 items leaves exactly 13 behind, so choosing 2 and choosing which 13 to leave out are the same count: 15 choose 2 = 15 choose 13 = 105. In the formula n! / (k! (n − k)!) the two factorials below just swap places.

How many ways are there if order matters?

P(15, 2) = 210. Each of the 105 combinations can be put in order in 2! = 2 ways, so permutations = combinations × k!: P(n, k) = n! / (n − k)!.