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18 Choose 2

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What is 18 choose 2?

Answer

18 choose 2 is

153

Order doesn't matter, no repeats: 153 ways to choose 2 of 18 items.

C(18, 2) = 18! / (2! · 16!)

= 153

If order matters (permutations): P(18, 2) = 18 × 17 = 306 — each combination in 2! = 2 orders.

Row 18 of Pascal's triangle
k	18 choose k
0	1
1	18
2	153
3	816
4	3,060
5	8,568
6	18,564
7	31,824
8	43,758
9	48,620
10	43,758
11	31,824
12	18,564
13	8,568
14	3,060
15	816
16	153
17	18
18	1

Row sum: 2¹⁸ = 262,144 — all subsets of 18 items.

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https://calculat.io/en/number/combinations/18--2

<a href="https://calculat.io/en/number/combinations/18--2">18 Choose 2: Combinations C(18, 2) - Calculatio</a>

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18 Choose 2: Combinations C(18, 2). 18 choose 2 = 153: the number of ways to choose 2 of 18 items when order doesn't matter and nothing repeats. C(18, 2) = (18 × 17) / (2 × 1).

https://calculat.io/en/number/combinations/18--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 18 choose 2?

18 choose 2 is 153

Why is 18 choose 2 the same as 18 choose 16?

Every way to choose 2 of 18 items leaves exactly 16 behind, so choosing 2 and choosing which 16 to leave out are the same count: 18 choose 2 = 18 choose 16 = 153. In the formula n! / (k! (n − k)!) the two factorials below just swap places.

How many ways are there if order matters?

P(18, 2) = 306. Each of the 153 combinations can be put in order in 2! = 2 ways, so permutations = combinations × k!: P(n, k) = n! / (n − k)!.

See Also