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From a group of M employees, N will be selected, at random, [#permalink]
07 Aug 2014, 13:58

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Difficulty:

55% (hard)

Question Stats:

43% (01:49) correct
57% (00:54) wrong based on 54 sessions

From a group of M employees, N will be selected, at random, to sit in a line of N chairs. There are absolutely no restrictions, either in the selection process nor in the order of seating — both are entirely random. What is the probability that the employee Andrew is seated somewhere to the right of employee Georgia? Statement #1: N = 15 Statement #2: N = M

Re: From a group of M employees, N will be selected, at random, [#permalink]
07 Aug 2014, 19:36

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mikemcgarry wrote:

From a group of M employees, N will be selected, at random, to sit in a line of N chairs. There are absolutely no restrictions, either in the selection process nor in the order of seating — both are entirely random. What is the probability that the employee Andrew is seated somewhere to the right of employee Georgia? Statement #1: N = 15 Statement #2: N = M

Statement 1: insufficient since we need to know M to solve for the probability Statement 2: N= M This turns to sit N people into N chairs For A is seated somewhere to the right of employee Georgia, we have (1+ 2+ 3+....+ (n-1)) possibilities To illustrate this, lets say 4 people sit in 4 chairs G in seat 1, A can be seated in 2,3, and 4 ( 3 possibilities) G in seat 2, A can be seated in 3,4 ( 2 possibilities) G in seat 3, A can be seated in 4 only (1 possibility) Sum = 1+ 2+ 3 After G and A are seated, there are (n-2)! possible combination ( for example, in this case, after G and A are seated, there are two seats left, we have 2! possible combination) The total possible combinations for N people in N chairs is N! The possible combinations that meet the requirement is (1+2+3+...(N-1))* (N-2)! = [(N-1)*N/2] * (N-2)! The probability = [(N-1)*N/2] * (N-2)!/N! = [(N-1)*N/2]/ [(N-1)*N] = 1/2 B is the answer. _________________

......................................................................... +1 Kudos please, if you like my post

Re: From a group of M employees, N will be selected, at random, [#permalink]
08 Aug 2014, 00:01

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Mmm:) That's easy.

(1) Not sufficient because we don't know M. (2) Sufficient. You need to arrange N people on N seats in such way that the employee Andrew is seated somewhere to the right of employee Georgia. The main idea that you have the same number of possibilities to put Andrew to the right of Georgia and to put him on the left. Just think that in each case when Andrew to the right of Georgia, you can switch their places. That's why, the probability is 1/2.

The correct answer is B. _________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos:)

Re: From a group of M employees, N will be selected, at random, [#permalink]
08 Aug 2014, 09:56

Expert's post

Dear smyarga, Good job! You took the very elegant solution to this.

Dear vad3tha, You found the answer via brute force calculations, but recognize that this approach can be costly, in terms of time & energy, on the real GMAT. If you want to achieve a truly elite GMAT score, part of what that requires is the ability to see the elegant solutions that involve few or no calculations. See: http://magoosh.com/gmat/2013/how-to-do- ... th-faster/

Re: From a group of M employees, N will be selected, at random, [#permalink]
10 Aug 2014, 18:39

mikemcgarry wrote:

Dear smyarga, Good job! You took the very elegant solution to this.

Dear vad3tha, You found the answer via brute force calculations, but recognize that this approach can be costly, in terms of time & energy, on the real GMAT. If you want to achieve a truly elite GMAT score, part of what that requires is the ability to see the elegant solutions that involve few or no calculations. See: http://magoosh.com/gmat/2013/how-to-do- ... th-faster/

Does all this make sense? Mike

I agree with you, also. This is not what I did when I first saw the problem. I answered by trying out: 4 people in 4 chairs and 5 people in 5 chairs. I came up with 1/2. I posted this solution afterward in case if ppl want to see the conceptual part the problem. Thanks for reminding me that. _________________

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gmatclubot

Re: From a group of M employees, N will be selected, at random,
[#permalink]
10 Aug 2014, 18:39