Stuck with combinations!

BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

This topic has expert replies

Post new topic Post Reply

Previous Topic Next Topic

Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 am; Location: Chennai, India; Thanked: 83 times; Followed by:5 members

Stuck with combinations!

by Uva@90 » Fri Oct 18, 2013 2:37 am

A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?
(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635

OA E

I proceeded the above problem as below,
Committee of 6 members can be chosen by
(8C2*5C4) + (8C3*5C3) = 700
I am struck here,
I unable to handle this condition
two of the men refuse to serve together
But I know something to be subtracted from 700.
So I went for E, only option less than 700.
But I would like to learn how to calculate the above red colored one.

Thanks in advance,

Regards,
Uva.

Known is a drop Unknown is an Ocean

Quote

Source: Beat The GMAT — Problem Solving | Fri Oct 18, 2013 2:37 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Fri Oct 18, 2013 3:27 am

Uva@90 wrote:A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?
(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635

OA E

I proceeded the above problem as below,
Committee of 6 members can be chosen by
(8C2*5C4) + (8C3*5C3) = 700
I am struck here,
I unable to handle this condition
two of the men refuse to serve together
But I know something to be subtracted from 700.
So I went for E, only option less than 700.
But I would like to learn how to calculate the above red colored one.

Thanks in advance,

Regards,
Uva.

Let A and B = the two men who cannot serve together.
From the 700 committees you have counted, we must subtract the BAD committees: those that include both A and B.

Bad case 1: 3 women and 1 man are chosen to serve with A and B
Number of ways to choose 3 women from 5 options = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of ways to choose 1 man from the 6 remaining men = 6C1 = 6.
To combine these options, we multiply:
10*6 = 60.

Bad case 2: 4 women are chosen to serve with A and B
Number of ways to choose 4 women from 5 options = (5*4*3*2)/(4*3*2*1) = 5.

Thus:
Good committees = 700-60-5 = 635.

Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Quote

Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 am; Location: Chennai, India; Thanked: 83 times; Followed by:5 members

by Uva@90 » Fri Oct 18, 2013 3:54 am

GMATGuruNY wrote:
Let A and B = the two men who cannot serve together.
From the 700 committees you have counted, we must subtract the BAD committees: those that include both A and B.

Bad case 1: 3 women and 1 man are chosen to serve with A and B
Number of ways to choose 3 women from 5 options = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of ways to choose 1 man from the 6 remaining men = 6C1 = 6.
To combine these options, we multiply:
10*6 = 60.

Bad case 2: 4 women are chosen to serve with A and B
Number of ways to choose 4 women from 5 options = (5*4*3*2)/(4*3*2*1) = 5.

Thus:
Good committees = 700-60-5 = 635.