Grouping question

[Eddie Gutia]

I came across a question given below.

A local card club will send 3 representatives to the national conference. If the local club has 8 members, how many different groups of representatives could the club send?

MGMAT answer: 8!/3!*5!

My doubt: If we want to solve this problem without using formula, below is my approach.

For the group 3 members, the first person can be any of the 8 members, 2nd person can be any of the 7 members and the third person can be any of the 6 members. So according to my interpretation the solution should be 8*7*6. According to MGMAT answer this is obviously. Can you please correct my approach?

[jlucero]

Compare your answer to the answer given:

8!/3!5! vs 8*7*6.

Note that you could make the second one into 8! by multiplying by 5*4*3*2*1, and then we can cancel that by multiplying by 5*4*3*2*1 on the bottom:

8!/3!5! vs 8*7*6*5*4*3*2*1/5*4*3*2*1

Cleaning this up we get:

8!/3!5! vs 8!/5!

So what did you forget to do? Divide by a 3!. Why is this important? Well think about your math- when you say 8 people can be selected first and then 7 are selected second, that means 56 combinations could be selected first and second. But we have some repeats. Selecting Amy first and Betty second is no different than selecting Betty first and Amy second. So we divide by 2! in this case to say that even though we can select people in 56 different ways, there are 2 different ways to select the same 2 people, so there are only 28 different combinations. So what would you need to do finish up your method? And why?

[Eddie Gutia]

Great explanation.. Thanks a lot..

[tim]

:)