BTGModeratorVI wrote: ↑

Mon Apr 13, 2020 3:38 pm

Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary employees are women, how many of the possible groups of 4 employees consist of 3 women and one man?

A. 22
B. 35
C. 56
D. 70
E. 105

Answer: D
Source: Official GMATPrep test

So, there are 7 men and 5 women and we have to choose 4 of them such that there are 3 women and 1 man.

Possible ways = 5C3*7C1 = 10*7 = 70

The correct answer: D

Hope this helps!

-Jay
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BTGModeratorVI wrote: ↑

Mon Apr 13, 2020 3:38 pm

Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary employees are women, how many of the possible groups of 4 employees consist of 3 women and one man?

A. 22
B. 35
C. 56
D. 70
E. 105

Answer: D
Source: Official GMATPrep test

We are asked to find the number of ways of choosing 3 women from a group of 5 women and 1 man from a group of 7 men.

Let’s first find the number of ways to choose 3 women from 5. Since the order of how the 3 women are chosen doesn’t matter, we use combinations:

5C3 = 5!/(3! x 2!) = (5 x 4 x 3)/3! = 60/6 = 10

Similarly, the number of ways to choose 1 man from a group of 7 men is:

7C1 = 7!/(1! x 6!) = 7

Finally, the number of ways to choose 3 women from 5 women and 1 man from 7 men is:

5C3 x 7C1 = 10 x 7 = 70

Answer: D

BTGModeratorVI wrote: ↑

Mon Apr 13, 2020 3:38 pm

Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary employees are women, how many of the possible groups of 4 employees consist of 3 women and one man?

A. 22
B. 35
C. 56
D. 70
E. 105

Answer: D
Source: Official GMATPrep test

Take the task of selecting the employees and break it into stages.

Stage 1: Select the 3 women
The order in which we select the women does not matter, so we can use combinations.
We can select 3 women from 5 women in 5C3 ways (= 10 ways)

Aside: If anyone is interested, we have a free video on calculating combinations (like 5C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Select the 1 man
There are 7 men, so we can complete this stage in 7 ways.

By the Fundamental Counting Principle (FCP), we can complete the two stages (and thus select the permanent employees) in (10)(7) ways ([spoiler]= 70 ways[/spoiler])

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch this free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent