Permutations

[MrBen5]

Hmmm, i need someone clever for this.
If i had 10 numbers (e.g. 40-49) and wanted to perm them in sets of 3 (e.g. 40,41,42)
How many permutations would it be?
cheers...

[dundurn]

Assuming they don't have to be consecutive there are 10 ways to pick the first number, 9 ways to pick the second and 8 ways to pick the third which gives 720 ways to pick all three. Assuming the order isn't important you divide this answer by 3x2x1 = 6 to get the number of combinations = 120

[MrBen5]

Cheers :)