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16 Jan 2021, 01:54
Kudos
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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
i solved like this:
total number of ways: 8!
favorable outcomes: 8C2*6!*2!
my answer is not coming
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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16 Jan 2021, 03:33
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mufcshank
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.
16 Jan 2021, 17:40
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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
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
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
