Anagrams

[ColinM713]

Is there a different way to solve this problem other than using an anagram grid for this combinations questions?

A student committee that must consist of 5 members is to be formed from a pool of 8 candidates. How many different committees are possible?
A. 5
B. 8
C. 40
D. 56
E. 336

Thanks.

[RonPurewal]

you can use the more generic theory that underlies that whole thing:

• there are 8 x 7 x 6 x 5 x 4 different ways to pick the five people in order.

• however, order isn't supposed to matter, so we need to eliminate repetitions. there are 5! possible rearrangements of any given set of 5 people, and all of these are going to appear, so every committee is actually repeated 5! times in the calculation above.

• so, the total number of committees is (8 x 7 x 6 x 5 x 4) / 5!

[RonPurewal]

... and you can also just count the possibilities.

it might be a bit much to list all the different committees of 5 people. but, specifying 5 people is equivalent to specifying the 3 people who are left off the committee.

so, this problem is exactly equivalent to the situation in which we're picking 3, rather than 5, members.

these can definitely be counted in 1-2 minutes (123, 124, 125, 126, ..., 678).