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A team of 6 cooks is chosen from 8 Men and 5 Women

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A team of 6 cooks is chosen from 8 Men and 5 Women

by johndoe88 » Sun Sep 01, 2013 6:08 pm
A team of 6 cooks is chosen from 8 Men and 5 Women. The team must have at least 2 men and at least 3 woen. How many ways can we form this team?

A) 140
B) 320
C) 560
D) 700
E) 840

Solution: We have two possibilities 2M 4F or 3M 3F
(8C2)(5C3)+(8C3)(5C3)=700

I am trying another method but I am not getting it correctly. So for The first scenario 1) 3M 4F
MMFFFF
We can use a permutation approach. For the first man we have 8 choices. Second man 7. Likewise for the females. 5 for the first....2 for the last. We get
(8*7)*(5*4*3*2)
Now here is what is confusing me. Since I have MMFFFF and order doesn't matter the number of ways to arrange 3M's & 4F's is 6!/(2!*4!). I know I have to divide by 2! and 4! because we have duplicates and order doesn't matter but what about the 6!??? Should we include or exclude it? The answer has it excluded, but why?? I feel without the 6! we don't take into account MFFFMF or FFMFMF.... etc. What am I doing wrong?
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by vinay1983 » Sun Sep 01, 2013 8:53 pm
This is how I think it is:

Out of 8 men at least 2 men have to be in the team of 6 cooks, so there can be 2 or 3 men, while no.of women can be 3 or 4
Case I

Here no repetition is allowed and order is not important

So, 8C2 * 5c4 = 28 * 5=140

OR Case II

3 Men and 3 Women

8C3 * 5C3
= 56 * 10
=560

Hence 140 or 560=140+560=700

What is the source and OA?
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by vipulgoyal » Sun Sep 01, 2013 9:06 pm
Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

8c2*5c4+8c3*5c3 = 700
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by Brent@GMATPrepNow » Sun Sep 01, 2013 9:21 pm
johndoe88 wrote:A team of 6 cooks is chosen from 8 Men and 5 Women. The team must have at least 2 men and at least 3 women. How many ways can we form this team?
A) 140
B) 320
C) 560
D) 700
E) 840
For those of you who might not understand some of the more abbreviated solutions, here's the full solution.

We must consider two cases:
Case a) select 2 men and 4 women
Case b) select 3 men and 3 women

Case a) select 2 men and 4 women
We can take the task of selecting the 6 cooks and break it into stages.

Stage 1: Select the 2 men
Since the order in which we select the men does not matter, we can use combinations.
We can select 2 men from 8 men in 8C2 ways (28 ways)

If anyone is interested, we have a free video on calculating combinations (like 8C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Select the 4 women
We can select 4 women from 5 women in 5C4 ways (5 ways)

By the Fundamental Counting Principle (FCP) we can complete the 2 stages (and thus select the 6 cooks) in (28)(5) ways
So there are 140 possibles teams with 2 men and 4 women


Case b) select 3 men and 3 women
We can take the task of selecting the 6 cooks and break it into stages.

Stage 1: Select the 3 men
We can select 3 men from 8 men in 8C3 ways (56 ways)

Stage 2: Select the 3 women
We can select 3 women from 5 women in 5C3 ways (10 ways)

By the Fundamental Counting Principle (FCP) we can complete the 2 stages (and thus select the 6 cooks) in (56)(10) ways
So there are 560 possible teams with 3 men and 3 women

So, the total number of teams = 140 + 560 = [spoiler]700 = D[/spoiler]

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by johndoe88 » Mon Sep 02, 2013 1:14 pm
I know how to solve this problem. My question is what is wrong with this following approach:


So for The first scenario 1) 3M 4F
MMFFFF
We can use a permutation approach. For the first man we have 8 choices. Second man 7. Likewise for the females. 5 for the first....2 for the last. We get
(8*7)*(5*4*3*2)
Now here is what is confusing me. Since I have MMFFFF and order doesn't matter the number of ways to arrange 3M's & 4F's is 6!/(2!*4!). I know I have to divide by 2! and 4! because we have duplicates and order doesn't matter but what about the 6!??? Should we include or exclude it? The answer has it excluded, but why?? I feel without the 6! we don't take into account MFFFMF or FFMFMF.... etc. What am I doing wrong?
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by [email protected] » Mon Sep 02, 2013 2:06 pm
Hi johndoe88,

With your approach: MMFFFF, you end up multiplying numbers together:

(8x7)(5x4x3x2)

Since you're trying to eliminate duplicates, you have to divide by 2! (to account for the men) and 4! (to account for the women).

The reason WHY you don't have to deal with the 6! is because it's already accounted for in the calculation. When you multiply numbers together, it doesn't matter what comes first...

So: (8x7)(5x4x3x2) is the same as (5x4x3x2)(8x7) and the same as any other order you can come up with. For example: (2x5x8x7x4x3).

In the end, you'd have: (56 x 120) / (2 x 24) = 560.

This is what you would get with the other approaches as well.

You'd probably be better served learning and practicing the "easier" approach. You've run into this issue before (attempting to do the math in a much harder way and getting stuck) and you want to minimize the likelihood of it happening again (especially on Test Day).

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by ganeshrkamath » Tue Sep 03, 2013 3:09 am
johndoe88 wrote:A team of 6 cooks is chosen from 8 Men and 5 Women. The team must have at least 2 men and at least 3 woen. How many ways can we form this team?

A) 140
B) 320
C) 560
D) 700
E) 840

Solution: We have two possibilities 2M 4F or 3M 3F
(8C2)(5C4)+(8C3)(5C3)=700

I am trying another method but I am not getting it correctly. So for The first scenario 1) 2M 4F
MMFFFF
We can use a permutation approach. For the first man we have 8 choices. Second man 7. Likewise for the females. 5 for the first....2 for the last. We get
(8*7)*(5*4*3*2)
Now here is what is confusing me. Since I have MMFFFF and order doesn't matter the number of ways to arrange 2M's & 4F's is 6!/(2!*4!). I know I have to divide by 2! and 4! because we have duplicates and order doesn't matter but what about the 6!??? Should we include or exclude it? The answer has it excluded, but why?? I feel without the 6! we don't take into account MFFFMF or FFMFMF.... etc. What am I doing wrong?
6!/(2!*4!) is the formula for number of ways of making two teams of 2 and 4 out of 6 people.
6! is the number of ways of arranging the 6 people.
2! and 4! eliminate the order of arrangement.

In the problem above, this is not the case.
You have already selected the people (8*7)(5*4*3*2)
Now you have to just eliminate the order of arrangement.
So you get: (8*7)/2! * (5*4*3*2)/4! = 8C2 * 5C4

Cheers
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