Anthony and Michael sit on the six-member

[seraphicgirl]

see
M will be on one of the 6 seats
if he is on seat no 1, A will either be on his sub committee or will not be (illustration 1 and 5)

if he is on seat no.2, A will either be on his sub comm or will not be (illustration 2 and 4)

if he is on seat no.3, A will either be on his sub comm or not.but when A is on his sub comm then it would already be like the one covered in illustration 1, so we leave one of these out.

hence 5 cases, out of which only in 2 will A be in the same sub com.

[jnelson0612]

Oh yes! DUH.

I think of it as:

1) Make two committees of three:
__ __ __
__ __ __

2) Place Anthony on one committee:

A __ __
__ __ __

3) There are now five empty places for Michael. Two of those are on Anthony's committee, thus the answer is 2/5.

It seems you were saying the same thing. Thanks!

[seraphicgirl]

yes! :)

[jnelson0612]

:-)

[AnkurR175]

Now if we want to find the number of subcommittees that include M, we have two spots remaining that can be filled with five Board members (A, B, C, D, or E). Using the Anagram approach, we can label "Y" if they are selected on the subcommittee, or "N" if they are not.

I still don't get it.
How can you have a total 10 possible ways of creating 2 subcommittees with 3 members each from a total of 6 people without having everyone among the 6 folks as part of either of the 2 subcommittees?
So technically, Michael (or for that matter anyone among the 6) will always be part of one or the other subcommittee.

The way I see it is how can 6 people be divided into 2 subcomms having 3 people each.........6!/ 3!*3! = 20.

[RonPurewal]

frankly, it seems pretty silly to use combinatorial methods on this problem at all.

obviously, using combinatorial methods isn't particularly "easy" or "straightforward" (look at this discussion) -- and, just as obviously, we can just MAKE AN EXHAUSTIVE LIST, which will contain a VERY small number of entries.

the question is:
what percent of all the possible subcommittees that include Michael also include Anthony?

so, if we can just list all of those... we can simply COUNT the ones that have anthony on them.

let's say the six people are M (michael), A (anthony), "1", "2", "3", and "4".
thenall of the committees that include Michael are
M A 1
M A 2
M A 3
M A 4
M 1 2
M 1 3
M 1 4
M 2 3
M 2 4
M 3 4

there are ten of these, of which exactly 4 contain anthony... so the requisite probability is 4 out of 10 = 2/5.

if the NUMBER OF POSSIBILITIES IS SMALL ... don't bother with combinatorial methods! nothing but grief, difficulty, and time-wasting to be had there.