There are twelve people on the council of doom. Five ultimate doomsters need to be chosen by the mega doom man. How many groups of five different ultimate doomsters can he have?
2007-05-10 08:03:13 · 6 answers · asked by dannythedolphin 2 in Education & Reference ➔ Homework Help
It's a combination of 12 people taken 5 at a time:
C = 12!/(7!*5!)
C = (12*11*10*9*8)/(5*4*3*2*1)
( I already canceled the 7!)
C = 95040/120
C = 792 combinations
.
2007-05-10 08:14:19 · answer #1 · answered by Robert L 7 · 0⤊ 1⤋
This is a simple combination:
C(12,5) = 12! / 5!*(12-5)!= 12*11*10*9*8/5*4*3*2*1 = 792 groups
2007-05-10 15:18:28 · answer #2 · answered by Cato_I 4 · 0⤊ 0⤋
None - what is a comitee!
If you mean committee then the answer is 12x11x10x9x8/ 5x4x3x2x1 = 792.
2007-05-10 16:51:16 · answer #3 · answered by Anonymous · 0⤊ 0⤋
use the 'nCr' button on your calculator. N is total number you are choosing from. r is number you want to pick. So 12C5= 792. Hope this helps.
2007-05-10 15:31:20 · answer #4 · answered by Rafa-No1 3 · 0⤊ 1⤋
It is an "n choose k" problem.
Formula for n choose k:
n!
--------
k!(n-k)1
For your problem:
n=12, k=5...
12!
-------- =
5!(12-5)!
12!
------ =
5!7!
12x11x10x9x8
-------------------- =
5x4x3x2x1
792
2007-05-10 15:14:05 · answer #5 · answered by George R 2 · 1⤊ 1⤋
792 (use permutations)
2007-05-10 15:12:00 · answer #6 · answered by Ana 4 · 0⤊ 2⤋