Combinatorics Problem

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quiimari: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Thu Nov 10, 2011 10:12 am

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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Source: Beat The GMAT — Problem Solving | Fri Nov 11, 2011 8:28 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

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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ArunangsuSahu: Master | Next Rank: 500 Posts; Posts: 382; Joined: Thu Mar 31, 2011 5:47 pm; Thanked: 15 times

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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Ananthakrishnan: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sat Sep 18, 2010 11:23 pm; Thanked: 1 times

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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shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

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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ArunangsuSahu: Master | Next Rank: 500 Posts; Posts: 382; Joined: Thu Mar 31, 2011 5:47 pm; Thanked: 15 times

by ArunangsuSahu » Sat Nov 12, 2011 5:35 am

@Anantha--U r dealing with duplicates

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