2007-01-26 11:36:40 · 5 answers · asked by annita p 2 in Science & Mathematics ➔ Mathematics
6C4 = (6!)/(4!(6 - 4)!)
6C4 = (6!)/(4! * 2!)
6C4 = 720/48
6C4 = 15
ANS : 15 combinations
2007-01-26 15:19:00 · answer #1 · answered by Sherman81 6 · 0⤊ 0⤋
Same as the number of combinations of 2 things, because each time you choose 2, there are 4 left behind, so every set of 2 corresponds exactly to a set of 4, and vice versa.
6C4 = 6C2 = 15
With so few, you can write them out explicitly:
Choose from A,B,C,D,E,F
AB, AC, AD, AE, AF
BC, BD BE, BF, CD
CE, CF,DE, DF, EF
2007-01-26 19:42:43 · answer #2 · answered by Hy 7 · 0⤊ 0⤋
(6*5*4*3)/(4*3*2*1)
2007-01-26 19:42:59 · answer #3 · answered by bruinfan 7 · 0⤊ 0⤋
question is not clear..if my undestanding is correct then the answer is 60 (6C4 = 6!(6-4)! / 4!)
2007-01-26 19:46:29 · answer #4 · answered by Anonymous · 0⤊ 0⤋
C(6,4)
=C(6,2)
= 15
2007-01-26 19:41:58 · answer #5 · answered by sahsjing 7 · 0⤊ 0⤋