Question

How many different 9-member boards of directors can be formed from a group of 14 candidates?

I tried 14)13)12)11)10)9)8/14 but I'm not getting the right answer.

Answer 1

14C9
\[14C9=\frac{14!}{9!5!}=\frac{14\times 13\times 12\times 11\times 10}{5\times 4\times 3\times 2}=you\ can\ calculate\]

Answer 2

Okay...so 'splain this to me if you would be so kind.

Answer 3

What don't you understand?

Answer 4

Why this is so different from what my book said :). Okay, for starters, why are there only 5 terms on top and 4 on the bottom? Why five and four?

Answer 5

Do you understand what a factorial is?
The formula for finding the number of combinations of n different things taken r at a time is
\[nCr=\frac{n!}{r!(n-r)!}\]

Answer 6

No, never got that handy dandy formula. But okay...so you used 5 terms on the top and bottom (the five came from 14-9) but there's only 4 terms on the bottom because you just dropped the times 1 which would have no effect...am i right?

Answer 7

\[\frac{14!}{9!}=14\times 13\times 12\times 11\times 10\]
and
\[5!=5\times 4\times 3\times 2\]
As you realise the times 1 is unnecessary.

Answer 8

Okay I think I get how to use it...I dont completely get the why and how but thats just a luxury I guess.

Answer 9

There is a good explanation of combinations and permutations and the necessary formulas here:
http://www.mathsisfun.com/combinatorics/combinations-permutations.html

Answer 10

Okay i'll check it out. thanks

Answer 11

You're welcome :)