SEARCH HOME
 Math Central Quandaries & Queries
 Question from Jackie: I would like to know all the combinations possible with 3 numbers in each combination, using the numbers 1-14. Thanks

Hi Jackie,

I am assuming that you can't repeat numbers.

The number of ways of choosing k things from n things is n!/[k! (n - k)!] where n! is n × (n - 1) × (n - 2) ×···× 2 × 1. Thus the number of ways of choosing 3 numbers from 14 numbers is

14!/[3! 11!] = 364.

I am not going to list them for you but I will show you a way to list them yourself. Think of the numbers from 1 to 14 as the first 14 letters in the alphabet and list all 3 letter words made from these letters, with no repeated letters, in alphabetical order. They start this way

1, 2, 3
1, 2, 4
1, 2, 5
.
.
.
1, 2, 14
1, 3, 4
1, 3, 5
.
.
1, 3, 14
.
.
.
1, 13, 14
2, 3, 4
2, 3, 5
.
.
.
2, 3, 14
3, 4, 5
.
.
.
12, 13, 14

I hope this helps,
Penny

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.