A three-person committee must be chosen from a group of

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Gmat_mission: Legendary Member; Posts: 1622; Joined: Thu Mar 01, 2018 7:22 am; Followed by:2 members

A three-person committee must be chosen from a group of

by Gmat_mission » Sun May 13, 2018 10:30 am

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A three-person committee must be chosen from a group of 7 professors and 10 graduate students. If at least one of the people on the committee must be a professor, how many different groups of people could be chosen for the committee?

A. 70
B. 560
C. 630
D. 1,260
E. 1,980

[spoiler]OA=B[/spoiler].

Could anyone give me some help here? Please. I would like to know how to solve this PS question. <i class="em em-sob"></i>

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Source: Beat The GMAT — Problem Solving | Sun May 13, 2018 10:30 am

Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pm; Thanked: 6 times; Followed by:5 members

by Vincen » Sun May 13, 2018 11:31 am

Gmat_mission wrote:A three-person committee must be chosen from a group of 7 professors and 10 graduate students. If at least one of the people on the committee must be a professor, how many different groups of people could be chosen for the committee?

A. 70
B. 560
C. 630
D. 1,260
E. 1,980

[spoiler]OA=B[/spoiler].

Could anyone give me some help here? Please. I would like to know how to solve this PS question. <i class="em em-sob"></i>

Hi gmat_mission.

Here is how I'd solve it.

A three-person committee ------------> this implies that we have to pick a total of 3 people.
7 professors -------------------------------> Group 1 of people.
10 graduate students --------------------> Group 2 of people.

If at least one of the people on the committee must be a professor --------> this implies that we could pick 1 professor, 2 professors or 3 professors.

How many different groups of people could be chosen for the committee?

Now, the number of ways of forming the committee is:

- If we pick only one professor and two students:
$$7\ C\ 1\cdot10\ C\ 2\ =\ \frac{7!}{6!}\cdot\frac{10!}{8!2!}=7\cdot45=315.$$

- If we pick two professors and one student:
$$7\ C\ 2\cdot10\ C\ 1\ =\ \frac{7!}{5!2!}\cdot\frac{10!}{9!}=21\cdot10=210.$$

- If we pick three professors and none students:
$$7\ C\ 3\cdot10\ C\ 0\ =\ \frac{7!}{4!3!}\cdot1=35.$$

Finally, we have to add the results, hence we get: 315+210+35=560.

Therefore, the correct answer is the option B.

I hope it helps.

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 am; Location: New York, NY; Thanked: 39 times; Followed by:22 members

by Jeff@TargetTestPrep » Wed May 16, 2018 10:11 am

Gmat_mission wrote:A three-person committee must be chosen from a group of 7 professors and 10 graduate students. If at least one of the people on the committee must be a professor, how many different groups of people could be chosen for the committee?

A. 70
B. 560
C. 630
D. 1,260
E. 1,980

There are 17C3 = 17!/(3! x 14!) = (17 x 16 x 15)/(3 x 2) = 17 x 8 x 5 = 680 ways to form a 3-person committee from 17 people if there are no restrictions on how the 3 people are chosen. However, we are given that the 3-person committee must have at least one professor. To determine the number of ways we can do that, we can use the fact that

Number of committees with at least one professor = Total number of committees - number of committees with no professor

We've determined that the total number of committees is 680. Let's determine the number of possible committees with no professor. That is, the committee must have all graduate students and there are 10C3 = 10!/(3! x 7!) = (10 x 9 x 8)/(3 x 2) = 10 x 3 x 4 = 120 ways to form such a committee. Therefore,

Number of committees with at least one professor = 680 - 120 = 560

Answer: B

Jeffrey Miller
Head of GMAT Instruction
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