Where am I going wrong in this Q?

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sunil_snath: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Wed Nov 18, 2009 8:41 am; Thanked: 3 times

Where am I going wrong in this Q?

by sunil_snath » Sun Jan 10, 2010 7:58 am

From 6 gentlemen and 4 ladies a committee of 5 is to be formed. In how many ways can this be done, if the committee is to include atleast one lady?

A 246
B 252
C 120
D 492

Ans: A

Here's my approach and I cant find anything wrong:

Choose 1 lady: 4C1 = 4
Choose 4 remaining from the 9 people left: 9C4

Total combinations: 4*9C4 = 544 which is way off target. Any ideas?

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Source: Beat The GMAT — Problem Solving | Sun Jan 10, 2010 7:58 am

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Sun Jan 10, 2010 8:52 am

sunil_snath wrote:From 6 gentlemen and 4 ladies a committee of 5 is to be formed. In how many ways can this be done, if the committee is to include atleast one lady?

A 246
B 252
C 120
D 492

Ans: A

Here's my approach and I cant find anything wrong:

Choose 1 lady: 4C1 = 4
Choose 4 remaining from the 9 people left: 9C4

Total combinations: 4*9C4 = 544 which is way off target. Any ideas?

Your method is:
Stage 1: choose one woman from the 4 women
Stage 2: choose 4 people from the remaining people

This method guarantees that we have at least one woman, but it also allows you to count some combinations more than once.
For example, here's one possible combination:
Stage 1: Choose woman A
Stage 2: Choose women B, C, D and man E

Another combination is
Stage 1: Choose woman B
Stage 2: Choose women A, C, D and man E

Notice that these two combinations are the same, but they count as different combinations using your method.

Most "at least" questions are handled more quickly using complements.
Here, the # of committees with at least 1 woman = the total number of committees - the # of committees with 0 women

Total number of committees = 10C5 = 252 (10 people - choose 5 of them)
# of committees with 0 women = 6C5 = 6 (6 men - choose 5 of them)

# of committees with at least 1 woman = 252 - 6 = 246

Brent Hanneson - Creator of GMATPrepNow.com