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20 Choose 20

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What is 20 choose 20?

Answer

20 choose 20 is

1

Order doesn't matter, no repeats: 1 way to choose 20 of 20 items.

There is exactly one way: take all.

C(20, 20) = 20! / (20! · 0!)

= 1

0! = 1 (an empty product), so 20! cancels out completely.

If order matters (permutations): P(20, 20) = 20 × 19 × 18 × … × 1 = 2,432,902,008,176,640,000 — each combination in 20! = 2,432,902,008,176,640,000 orders.

Row 20 of Pascal's triangle
k	20 choose k
0	1
1	20
2	190
3	1,140
4	4,845
5	15,504
6	38,760
7	77,520
8	125,970
9	167,960
10	184,756
11	167,960
12	125,970
13	77,520
14	38,760
15	15,504
16	4,845
17	1,140
18	190
19	20
20	1

Row sum: 2²⁰ = 1,048,576 — all subsets of 20 items.

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See Also

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https://calculat.io/en/number/combinations/20--20

<a href="https://calculat.io/en/number/combinations/20--20">20 Choose 20: Combinations C(20, 20) - Calculatio</a>

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20 Choose 20: Combinations C(20, 20). 20 choose 20 = 1: the number of ways to choose 20 of 20 items when order doesn't matter and nothing repeats.

https://calculat.io/en/number/combinations/20--20/generated.jpg

Recalculate with different values ↑

FAQ

What is 20 choose 20?

20 choose 20 is 1

Why is 20 choose 20 the same as 20 choose 0?

Every way to choose 20 of 20 items leaves exactly 0 behind, so choosing 20 and choosing which 0 to leave out are the same count: 20 choose 20 = 20 choose 0 = 1. In the formula n! / (k! (n − k)!) the two factorials below just swap places.

How many ways are there if order matters?

P(20, 20) = 2,432,902,008,176,640,000. The only combination — all 20 items — can be put in order in 20! = 2,432,902,008,176,640,000 ways, so permutations = combinations × k!: P(n, k) = n! / (n − k)!.

See Also