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Combinatorics Problem

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by quiimari » Fri Nov 11, 2011 8:28 am
A five member committee is to be selected from among four Math teachers and five English teachers. In how many different ways can the committee be formed if the committee must contain at least three Math teachers?

A 40
B 45
C 60
D 21
E 126
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by GMATGuruNY » Fri Nov 11, 2011 8:37 am
quiimari wrote:A five member committee is to be selected from among four Math teachers and five English teachers. In how many different ways can the committee be formed if the committee must contain at least three Math teachers?

A 40
B 45
C 60
D 21
E 126
Case 1: 3 math teachers, 2 English teachers
Number of ways to choose 3 math teachers from 4 choices = 4C3 = 4.
Number of ways to choose 2 English teachers from 5 choices = 5C2 = 10.
To combine these options, we multiply:
4*10 = 40.

Case 2: 4 math teachers, 1 English teacher
Number of ways to choose 4 math teachers from 4 choices = 4C4 = 1.
Number of ways to choose 1 English teacher from 5 choices = 5.
To combine these options, we multiply:
1*5 = 5.

Total possible committees = 40+5 = 45.

The correct answer is B.
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by ArunangsuSahu » Fri Nov 11, 2011 9:07 pm
1)4C3*5C2 = 40
2)4C4*5C1 = 5

Total 40+5=45
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by Ananthakrishnan » Fri Nov 11, 2011 9:28 pm
I solved it this way.

3 math teachers can be selected in 4 (4C3) ways. Remaining 2 teachers can be selected from the remaining total available 6 teachers in 6C2 = 15 ways. So total number of ways is 4 x 15 = 60 ways.

I know that this answer is wrong. But, where am i going wrong?
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by shankar.ashwin » Fri Nov 11, 2011 10:15 pm
You don't want to mix up the 2 groups for you get a lot of duplicates.

Imagine (1,2,3,4) in one group and (5,6,7,8,9) in the other.

Now when you do 4C3 in the first group, you pick 3 from the 4 say (1,2,3) or ( 1,3,4) etc.

Now when you combine both the groups you can get combinations such as (4,5) (5,6) (2,5) etc.

Finally all we need is set of 5 numbers (1,2,3,4,5) (1,2,4,5,2) is the same. You would get 15 such duplicates.
Ananthakrishnan wrote:I solved it this way.

3 math teachers can be selected in 4 (4C3) ways. Remaining 2 teachers can be selected from the remaining total available 6 teachers in 6C2 = 15 ways. So total number of ways is 4 x 15 = 60 ways.

I know that this answer is wrong. But, where am i going wrong?
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by ArunangsuSahu » Sat Nov 12, 2011 5:35 am
@Anantha--U r dealing with duplicates
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