# Combinations: Two Approaches

If a team of 4 members is to be selected from 5 males and 4 females consisting
atleast 1 male and 1 female, then the number of ways of selecting the team is

a) 1680
b) 1200
c) 120
d) 80


## Standard Approach:

1 M 3 F = 5C1*4C3 = 20
3 M 1 F = 4C1*5C3 = 40
2 M 2 F = 5C2*4C2 = 60
=>120


## 2nd Approach:

1M can be selected in 5C1 = 5ways
1F can be selected in 4C1 = 4ways
Now we need to select 2 people out of (9-2=7) people
This can be done in 7C2 ways = 21 ways
Therefore,total number of ways = 5*4*21 = 420 ways
I know this is not the correct approach. Can anyone please
explain what's wrong with this approach?

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The problem with the second approach is that it counts each team more than once. Consider the team whose members are Anne, Bill, Gwen, and Tom, for instance: I claim that it gets counted four times.

1. First we choose Bill and Anne, and then we choose two of the remaining seven and happen to get Gwen and Tom.
2. First we choose Bill and Gwen, and then we choose two of the remaining seven and happen to get Anne and Tom.
3. First we choose Tom and Anne, and then we choose two of the remaining seven and happen to get Gwen and Bill.
4. First we choose Tom and Gwen, and then we choose two of the remaining seven and happen to get Anne and Bill.

In fact, every team with two males and two females is counted four times. You can check that each team with one male and three females or the reverse is counted three times.

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Ah I get it. Thank you. –  segmentation_fault Feb 20 '13 at 10:03
@segmentation_fault: You’re welcome. –  Brian M. Scott Feb 20 '13 at 10:03

The first approach is correct.

The mistake of the second approach is, that you are counting some combinations multiple times. You calculate the number of combinations, of which the order of the first man and woman matters.

For instance, look at this scenario. Your team consists of Man 1, Man 5, Woman 2, and Woman 3. You count this case four times, once during selecting Man 1 first (5C1) and Woman 2 second (4C1), once during selecting Man 1 first (5C1) and Woman 3 second (4C1), once during selecting Man 2 first (5C1) and Woman 2 second (4C1), and once during selecting Man 2 first (5C1) and Woman 3 second (4C1).

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Thanks a lot. I wish I could accept both the answers. –  segmentation_fault Feb 20 '13 at 10:03