Quick Math Combination Question

[FabienO]

Okay, 127 must be the right answer, if order doesn't matter.

If you only want one number, then you have 7C1 (ie: 7) to choose from.

If you want two numbers, you have 7C2=21 to choose from.

If you want three numbers, you have 7C3=35 to choose from.

7C4=35
7C5=21
7C6=7
7C7=1

1+7+21+35+35+21+7 = 127

[This assumes that order does not matter. Ie - if you pick number 5 and 2, that's the same as 2 and 5]

Awesome, cheers

 

[joxang]

Combinations of 1 = 7
Combinations of 2 = [7!] / [2! * (7-2)!]
Combinations of 3 = [7!] / [3! * (7-3)!]
etc

 

[Greebo]

Permutations would be 7! which is indeed 5040 However, the op has not stated how many out of the 7 numbers he is picking. Strictly speaking there is only one combination of 7 from 7 numbers

So we have

7 from 7 = 1
6 from 7 = 7
5 from 7 = 21
4 from 7 = 35
3 from 7 = 35
2 from 7 = 21

Total combinations = 120

n_C_x represents the number of combinations of n items taken x at a time.
n_P_x represents the number of permutations of n items taken x at a time.

n_C_x = n_P_x / x! and n_P_x = n!/(n-x)! therefore n_C_x = n!/(n-x)!x!

edit:I assume that you were missing out single numbers for some odd reason...............

 

[Freakbro]

also sentences shouldn't start with and or but.

That rule does not hold any longer, well it was never a rule just convention. It is perfectly fine to start sentences with a conjunction nowadays.

 

[Scon]

TheCenturion is correct

7! would imply you have to pick 7 numbers, and that order has no speficiation.

7! = 7*6*5*4*3*2*1

7 ways to pick first number
6 ways to pick second number
5 ""
4 ""
3 ""
2 ""
1 ""