Maths Question

[Kerplunk]

A long time since I did Maths at school...

If I have the numbers 1 -12, how many different combinations of 6 numbers can I have, order not important?

[mxyzptlk]

I make it 924

[Kerplunk]

Ok thanks, I have just found the formula.

C(12,6) = 12! / (6!) * (12 - 6)!

You can actually do this in excel (=combin(12,6))

[jenstar]

Unless you're allowed to use each number more than once, in which case it's 12^6.

[Jay70]

Isn't it 12 / 6 = 2  groups of 6 No

each group has 6x6 = 36 combinations

so the 2 groups have combined combinatios of 36 x 36 = 1296 total combinations?

[Hamish]

Err Kerplunk I make that 12! then...... I think it should be 12!/6!

Maths - 12 poss for the first, x 11 poss for the second etc. to 7 poss. ie 12x11x10x9x8x7.

[Loosehead]

It's the classic combination:

(12x11x10x9x8x7)/(1x2x3x4x5x6) = 924

Use the same in betting to work out how many doubles or trebles etc

eg how many trebles in a heinz:

(6x5x4)/(1x2x3) = 120/6 = 20

[Space]

kerplunk probably meant
12! / ( 6! * (12-6)! ) 
Hamish, you calculated the number of combinations, order being important. 
Since order is not important, you have to divide that result by 6! which is the number of possible ways the combinations can be ordered.

[Hamish]

Yes indeed, I thought it didn't look quite right. Long time since I had to do this sort of maths!