GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Apr 2019, 09:47

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

M18-23

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 54369

Show Tags

16 Sep 2014, 01:04
00:00

Difficulty:

5% (low)

Question Stats:

91% (00:56) correct 9% (01:50) wrong based on 177 sessions

HideShow timer Statistics

10 engineers and 3 designers have volunteered to work on a project. If the project team must consist of 3 engineers and 2 designers, in how many different ways can the team be formed?

A. 360
B. 400
C. 420
D. 450
E. 600

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 54369

Show Tags

16 Sep 2014, 01:04
1
Official Solution:

10 engineers and 3 designers have volunteered to work on a project. If the project team must consist of 3 engineers and 2 designers, in how many different ways can the team be formed?

A. 360
B. 400
C. 420
D. 450
E. 600

Engineers can be selected in $$C_{10}^{3} = \frac{10!}{7! 3!} = 120$$ different ways. Designers can be selected in $$C_{3}^{2} = \frac{3!}{2! 1!} = 3$$ different ways. The answer to the question is therefore $$120*3 = 360.$$

_________________
Intern
Joined: 10 May 2017
Posts: 27

Show Tags

30 Jun 2017, 22:55
I agree with the explanation.

One of the important rule learnt is AND means multiplication and OR means addition.

This is not a probability question.

It is asking ways to arrange the engineers and designers and the order doesn't matter. A simple anagram question.
Intern
Joined: 23 May 2017
Posts: 3

Show Tags

04 Jul 2017, 05:32
I had a simple question, the problem says how many different ways, order doesn't matter, shouldn't we be using permutation than combination ? I think it should be 10P3 x 3P2
Math Expert
Joined: 02 Sep 2009
Posts: 54369

Show Tags

04 Jul 2017, 05:39
Ganeshsrinivasan wrote:
I had a simple question, the problem says how many different ways, order doesn't matter, shouldn't we be using permutation than combination ? I think it should be 10P3 x 3P2

Yes, order does not matter and that's why we are using combinations and not permutation. 10C3 gives different unordered groups of 3 out of 10 and 3C2 gives different unordered groups of 2 out of 3. Multiplying those two gives the answer.
_________________
Re: M18-23   [#permalink] 04 Jul 2017, 05:39
Display posts from previous: Sort by

M18-23

Moderators: chetan2u, 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®.