Feedback form

15 Choose 14

Combination Calculator

What is 15 choose 14?

Answer

15 choose 14 is

15

Order doesn't matter, no repeats: 15 ways to choose 14 of 15 items.

Choosing 14 of 15 is the same as leaving 1 out: C(15, 14) = C(15, 1).

C(15, 14) = 15! / (14! · 1!)

= C(15, 1)

= 15

If order matters (permutations): P(15, 14) = 15 × 14 × 13 × … × 2 = 1,307,674,368,000 — each combination in 14! = 87,178,291,200 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--14

<a href="https://calculat.io/en/number/combinations/15--14">15 Choose 14: Combinations C(15, 14) - Calculatio</a>

Facebook

Twitter

Telegram

WhatsApp

Viber

Email

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

https://calculat.io/en/number/combinations/15--14/generated.jpg

n choose 14 for other n

n choose 14	Answer
14 choose 14	1
15 choose 14	15
16 choose 14	120
17 choose 14	680
18 choose 14	3,060
19 choose 14	11,628
20 choose 14	38,760

Recalculate with different values ↑

FAQ

What is 15 choose 14?

15 choose 14 is 15

Why is 15 choose 14 the same as 15 choose 1?

Every way to choose 14 of 15 items leaves exactly 1 behind, so choosing 14 and choosing which one to leave out are the same count: 15 choose 14 = 15 choose 1 = 15. In the formula n! / (k! (n − k)!) the two factorials below just swap places.

How many ways are there if order matters?

P(15, 14) = 1,307,674,368,000. Each of the 15 combinations can be put in order in 14! = 87,178,291,200 ways, so permutations = combinations × k!: P(n, k) = n! / (n − k)!.