How many ways can 5 starters be selected from a basketball team of 12?
show ALL work, steps, and how you got the answer. thanks
2006-12-10 02:16:11 · 3 answers · asked by ÄÐЦÇT¦ÖÑ™ 4 in Science & Mathematics ➔ Mathematics
thats not what i'm looking for Daryl.
I'm looking for the ! rule.
2006-12-10 02:24:37 · update #1
A way to do it is combinations:
(12) = 12!/
( 5) = 5!(12-5)!
You divide those two
= 792
where 12! = 1*2*3*4*5*6*7*8*9*10*11*12
where 5! = 1*2*3*4*5
where (12-5)! = 7! = 1*2*3*4*5*6*7
so when you do the division 12!/7!5! = 8*9*10*11*12/1*2*3*4*5
which is 792
2006-12-10 02:23:38 · answer #1 · answered by Modus Operandi 6 · 0⤊ 0⤋
12*11*10*9*8 is your answer, or 95,040. At first, you have 12 players to randomly select from, hence the 12. Then, once he's out, 11 remain. There are 11 different ways to do this for each of the first possible 12 players. This pattern goes on til there are 8 players left.
2006-12-10 10:21:27 · answer #2 · answered by Anonymous · 0⤊ 1⤋
combinations is nCr
12C5
=12*11*10*9*8/1*2*3*4*5
=792
2006-12-10 11:13:09 · answer #3 · answered by raj 7 · 0⤊ 0⤋