Why doesn't (8c1)(13c2) work for choosing a team with at least one woman?

[gtfitzpatrick]

Homework Statement

A club has only 8 women and 6 men as members. A team of 3 is to be chosen to represent the club. In how many ways can this be done if there is to be at least one woman on the team.

Homework Equations

The Attempt at a Solution

I can do this 2 ways,
first 1w2m + 2w1m + 3m0m => (8c1)(6c2) + (8c2)(6c1) + (8c3)(6c0) =344
or
total- all men => (14c3) - (6c3) = 344

But the problem i can't get my head around is why this way doesn't work: (8c1)(13c2)
can some one please throw some light on this for me[/B]

 

[andrewkirk]

But the problem i can't get my head around is why this way doesn't work: (8c1)(13c2)

Because you are putting a partial order on the the three people selected. You select one, and then you select a pair to make up three. So some sets of three are counted twice. For instance, the case where you first pick Laxmi, then pick Mae and Brunhilde (together) will be treated as different from the one where you first pick Mae, then Laxmi and Brunhilde (together). But as a combination, they are just one instance, not two.

 

[Charles Link]

In your second way, you have 8 choices for lady number 1, and you select two more from the remaining 13 people. Some of the groups you form in this way will be duplicated= e,g. Suppose lady 1 is your first choice, and suppose lady 5 and lady 6 are the remaining two choices (from C(13 2)). Alternatively, suppose lady 5 is now the lady designated out of the C(8 1). Once again you will find a case where she is paired with a C(13 2) that happens to be ladies 1 and 6. This is a simple example, but there are numerous others=the result is the number C(8 1) C(13 2) is incorrect and double counts numerous times.