question

Question

If 8 students are to be seated in a row of 8 seats, what is the probability that two particular students are seated at the two extreme ends of the row?

i solved like this:

total number of ways: 8!

favorable outcomes: 8C2*6!*2!

my answer is not coming

mSKR · Answer

You are doing some small mistake and mixing 2 problems together , thats why your answer will be 1
The reason is: at one end u r selecting 2 people and these people can be anyone . and others are rearranging accordingly.
ABCDEFGH
so what u r doing is 
at 2 extreme ends : there can be A and B or A can be H or H or G-means at 2 ends there can be anyone according to 8c2 and then u r arraing 6 
in this case, answer would be always 1

Split it into 2 parts:
1. Fix the particular people at 2 ends . Example 2 females sit at corners and other 6 males sit in corner
In this way, 2 *6!/8! should be your answer.

2. Select 2 people who may sit at corner and rest doesnt matter how do they sit
8c2/8!

When u add 1 and 2 -answer will be 1

I hope it helps.

EMPOWERgmatRichC · Answer

Hi mufcshank,

To start, when posting a subject-specific question, you should post the ENTIRE prompt (including the 5 answer choices) in the appropriate sub-forum. For example, the Problem Solving (PS) forum is here:

problem-solving-ps-140/

With this question, you are correct that the total number of possible arrangements is 8!

Since we want 2 specific people (out of the 8 total people) on the 'ends' of the row, there are 2 possible outcomes that 'fit' what we want. If we refer to the two people as A and B, then we're looking for either:

A _ _ _ _ _ _ B

or

B _ _ _ _ _ _ A

with the other 6 people arranged in the 6 'middle' seats. Thus, the total number of outcomes that we're looking for is 2(6!).

The probability of that type of outcome happening is 2(6!)/8! = 2/(8)(7) = 2/56 = 1/28

GMAT assassins aren't born, they're made,
Rich

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