are possible?
2007-11-30 08:20:28 · 2 answers · asked by DuckyWucky 3 in Science & Mathematics ➔ Mathematics
From the 6 children we need to pick the 3 children that will talk to their parents:
6 choose 3 = 20
C(n, k) = n! / k! (n-k)!
C(6, 3) = 6! / 3! 3!
C(6, 3) = 720 / 36
C(6, 3) = 20
This is equivalent to:
6 x 5 x 4
------------ = 20
.... 3!
Greg, Peter, Bobby
Greg, Peter, Marcia
Greg, Peter, Jan
Greg, Peter, Cindy
Greg, Bobby, Marcia
Greg, Bobby, Jan
Greg, Bobby, Cindy
Greg, Marcia, Jan
Greg, Marcia, Cindy
Greg, Jan, Cindy
Peter, Bobby, Marcia
Peter, Bobby, Jan
Peter, Bobby, Cindy
Peter, Marcia, Jan
Peter, Marcia, Cindy
Peter, Jan, Cindy
Bobby, Marcia, Jan
Bobby, Marcia, Cindy
Bobby, Jan, Cindy
Marcia, Jan, Cindy
2007-11-30 08:29:20 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
For the first one of the bunch there are 1 out of 6 choices, for the 2nd 1 of 5 and for the last 1 of 4 i.e. 6 x 5 x 4 = 120 but each of this group could be the same ones in a different order so you need to divide by 1 x 2 x 3 = 6 so the answer is 20 ways
2007-11-30 08:29:55 · answer #2 · answered by Anonymous · 0⤊ 0⤋