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GMAT Combinations Problem

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GMAT Combinations Problem

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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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
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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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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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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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