GMAT Combinations Problem

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saleem.kh: Junior | Next Rank: 30 Posts; Posts: 25; Joined: Sat Jul 10, 2010 1:44 amFollowed by: 1 member; Location: Kuwait; Thanked: 23 times

by saleem.kh » Mon Jun 27, 2011 5:24 am

Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?

A) 28
B) 46
C) 55
D) 63
E) 70

I read this question on KAPLAN blog (https://j.mp/lElkkt) but not able to understand properly. Why to take 6! and adjusting divisors instead of 8! and adjusting divisors accordingly. Is there any other best way to solve it? I will be really thankful.

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Source: Beat The GMAT — Problem Solving | Mon Jun 27, 2011 5:24 am

Frankenstein: Legendary Member; Posts: 1448; Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members; Location: India; Thanked: 375 times

by Frankenstein » Mon Jun 27, 2011 5:29 am

Hi,
4 people can be selected from 8 in 8C4 ways = 70
Number of ways in which four can be selected such that neither Jane nor Thomas is selected is nothing but picking 4 from the remaining 6 in 6C4 ways = 15

So, number of ways for at least one of either Jane or Thomas is to be selected = 70-15 = 55

Hence, C

Cheers!

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diebeatsthegmat: Legendary Member; Posts: 1119; Joined: Fri May 07, 2010 8:50 amFollowed by: 3 members; Thanked: 29 times

by diebeatsthegmat » Mon Jun 27, 2011 6:40 pm

or you can choose jane or thomas first and choose the other people in 6 people left
6!/3!3!=20
so there are 20* 2= 40 ways to choose or jane or thomas
choose both jane and thomas and choose 2 from 6 people left
there are 15 days
totally is 55

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saleem.kh: Junior | Next Rank: 30 Posts; Posts: 25; Joined: Sat Jul 10, 2010 1:44 amFollowed by: 1 member; Location: Kuwait; Thanked: 23 times

by saleem.kh » Mon Jun 27, 2011 9:36 pm

Thank you so much dears. Your both styles of calculations are perfect to save times.

Regards,

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amit2k9: Master | Next Rank: 500 Posts; Posts: 461; Joined: Tue May 10, 2011 9:09 amFollowed by: 3 members; Location: pune; Thanked: 36 times

by amit2k9 » Tue Jun 28, 2011 3:35 am

8c4-6c4 = 55

6c4 for not selecting J and T.
8c4 for total combinations.

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