It is currently 18 Nov 2017, 18:48

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# maths 08 question #4 doubt

Author Message
Intern
Joined: 12 Jun 2009
Posts: 15

Kudos [?]: [0], given: 0

maths 08 question #4 doubt [#permalink]

### Show Tags

28 Oct 2009, 10:06
If 4 women and 6 men work in the accounting department, in how many ways can a committee of 3 be formed if it has to include at least one woman?

The official solution to the above problem is

Consider an unconstrained version of the question: in how many ways can a committee of 3 be formed? The answer is 10c3. From this figure we have to subtract the number of committees that consist entirely of men: 6c3 The final answer is 10c3-6c3=120-20=100.

However my doubt is why cant we solve it using below

no of ways=4c1*6c2+4c2*6c1+4c3 (sum of 3 diff cases)
1st case--->select 1 women out of 4 and 2 men out of 6 to make a team of 3
2nd case--->select 2 w and 1 m to make a team of 3
3rd case--->select all 3 members from 4 available women

Kudos [?]: [0], given: 0

SVP
Joined: 29 Aug 2007
Posts: 2471

Kudos [?]: 855 [0], given: 19

Re: maths 08 question #4 doubt [#permalink]

### Show Tags

28 Oct 2009, 14:27
ggmatt wrote:
If 4 women and 6 men work in the accounting department, in how many ways can a committee of 3 be formed if it has to include at least one woman?

The official solution to the above problem is

Consider an unconstrained version of the question: in how many ways can a committee of 3 be formed? The answer is 10c3. From this figure we have to subtract the number of committees that consist entirely of men: 6c3 The final answer is 10c3-6c3=120-20=100.

However my doubt is why cant we solve it using below

no of ways = 4c1*6c2 + 4c2*6c1 + 4c3 (sum of 3 diff cases)
1st case--->select 1 women out of 4 and 2 men out of 6 to make a team of 3
2nd case--->select 2 w and 1 m to make a team of 3
3rd case--->select all 3 members from 4 available women

You can do that way too as both are same.

1. 10c3 - 6c3 = 100
2. (4c1 x 6c2) + (4c2 x 6c1) + 4c3 = 4x15 + 6x6 + 4 = 60 + 36 + 4 = 100
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Kudos [?]: 855 [0], given: 19

Re: maths 08 question #4 doubt   [#permalink] 28 Oct 2009, 14:27
Display posts from previous: Sort by

# maths 08 question #4 doubt

Moderator: Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.