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10 Choose 9

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What is 10 choose 9?

Answer

10 choose 9 is

10

Order doesn't matter, no repeats: 10 ways to choose 9 of 10 items.

Choosing 9 of 10 is the same as leaving 1 out: C(10, 9) = C(10, 1).

C(10, 9) = 10! / (9! · 1!)

= C(10, 1)

= 10

If order matters (permutations): P(10, 9) = 10 × 9 × 8 × … × 2 = 3,628,800 — each combination in 9! = 362,880 orders.

Row 10 of Pascal's triangle
k	10 choose k
0	1
1	10
2	45
3	120
4	210
5	252
6	210
7	120
8	45
9	10
10	1

Row sum: 2¹⁰ = 1,024 — all subsets of 10 items.

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See Also

Random Number Generator - Pick a random whole number between two bounds
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https://calculat.io/en/number/combinations/10--9

<a href="https://calculat.io/en/number/combinations/10--9">10 Choose 9: Combinations C(10, 9) - Calculatio</a>

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10 Choose 9: Combinations C(10, 9). 10 choose 9 = 10: the number of ways to choose 9 of 10 items when order doesn't matter and nothing repeats. C(10, 9) = 10 / 1.

https://calculat.io/en/number/combinations/10--9/generated.jpg

n choose 9 for other n

n choose 9	Answer
9 choose 9	1
10 choose 9	10
11 choose 9	55
12 choose 9	220
13 choose 9	715
14 choose 9	2,002
15 choose 9	5,005
16 choose 9	11,440
17 choose 9	24,310
18 choose 9	48,620
19 choose 9	92,378
20 choose 9	167,960

Recalculate with different values ↑

FAQ

What is 10 choose 9?

10 choose 9 is 10

Why is 10 choose 9 the same as 10 choose 1?

Every way to choose 9 of 10 items leaves exactly 1 behind, so choosing 9 and choosing which one to leave out are the same count: 10 choose 9 = 10 choose 1 = 10. In the formula n! / (k! (n − k)!) the two factorials below just swap places.

How many ways are there if order matters?

P(10, 9) = 3,628,800. Each of the 10 combinations can be put in order in 9! = 362,880 ways, so permutations = combinations × k!: P(n, k) = n! / (n − k)!.