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# GMAT Club Hardest Questions: Probability and Combinatorics

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GMAT Club Hardest Questions: Probability and Combinatorics [#permalink]

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26 Mar 2011, 06:17
I was just working on the set of probability and combinatorics questions from the GMAT Club hardest questions set. I have a question about #13:

A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?

Does the order of selection matter here? The official answer states: Look for any of the three variants: Tom-notMary-notMary, notMary-Tom-notMary, notMary-notMary-Tom. But if we are just choosing 3 people, why does the "variants" matter. I am having difficulty understanding when permutations matter and when they do not. I was thinking that Tom must be on the committee, that leaves 2 spots available. Mary can't be on the committee, so that leaves 4 available people to choose from. Hence I did, 4 choose 2 as the number of ways to select the committee. The total number of ways to select the committee of 3 without restrictions would then be 6 choose 3.

Is my thinking way off?
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Re: GMAT Club Hardest Questions: Probability and Combinatorics [#permalink]

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26 Mar 2011, 08:22
mbizhtk wrote:
I was just working on the set of probability and combinatorics questions from the GMAT Club hardest questions set. I have a question about #13:

A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?

Does the order of selection matter here? The official answer states: Look for any of the three variants: Tom-notMary-notMary, notMary-Tom-notMary, notMary-notMary-Tom. But if we are just choosing 3 people, why does the "variants" matter. I am having difficulty understanding when permutations matter and when they do not. I was thinking that Tom must be on the committee, that leaves 2 spots available. Mary can't be on the committee, so that leaves 4 available people to choose from. Hence I did, 4 choose 2 as the number of ways to select the committee. The total number of ways to select the committee of 3 without restrictions would then be 6 choose 3.

Is my thinking way off?

Both methods would be correct. Yours faster, elegant and less hairy.

From the probability approach;

T & NM & NM
OR
NM & T & NM
OR
NM & NM & T

$$\frac{1}{6}*\frac{4}{5}*\frac{3}{4}+\frac{4}{6}*\frac{1}{5}*\frac{3}{4}+\frac{4}{6}*\frac{3}{5}*\frac{1}{4}$$ $$=\frac{3}{10}$$

$$P=\frac{C^{4}_{2}}{C^{6}_{3}} = \frac{6}{20} = \frac{3}{10}$$
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Re: GMAT Club Hardest Questions: Probability and Combinatorics   [#permalink] 26 Mar 2011, 08:22
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# GMAT Club Hardest Questions: Probability and Combinatorics

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