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kevincan
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sgrover
(E)

Lets for simplicity, assume the names as a, b, c, d and e (as there would always be a way of sorting them by names as the names are distinct)..

Total permutatons possible of the name ordering: 5! = 120

Now, lets find the favorable space,
Consider the sequence abcde

We need to shuffle two alphabets in the above sequence to yeild what we want.
choose two alphabets from the above sequence (so that we can swap)
5C2 = 10
prob = 10/120 = 1/12 = 0.08


oops.. read 5 ppl by mistake..
for 6 ppl, ans. shd be

6C2/6! = 15/720 = 0.02 (B)
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sgrover
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(E)

Lets for simplicity, assume the names as a, b, c, d and e (as there would always be a way of sorting them by names as the names are distinct)..

Total permutatons possible of the name ordering: 5! = 120

Now, lets find the favorable space,
Consider the sequence abcde

We need to shuffle two alphabets in the above sequence to yeild what we want.
choose two alphabets from the above sequence (so that we can swap)
5C2 = 10
prob = 10/120 = 1/12 = 0.08

oops.. read 5 ppl by mistake..
for 6 ppl, ans. shd be

6C2/6! = 15/720 = 0.02 (B)


Excellent thinking, sgrover, I declare this the OA and the OE!

I'd like comments on whether these questions are helpful. There is never an excess of difficult questions, and people shouldn't feel discouraged if the questions are hard at first glance. Just try to learn techniques to apply them to other questions. That's what I do!
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kevincan,

what is the source of these questions.. these are indeed tricky..
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I'm the GMAT trainer at Kaplan, Madrid, Spain and I make these questions up for my students. (They like them). Do people mind if they are not OG or GMATPrep questions? I know they aren't easy, but I really like the thinking/explanations people are providing and I think people can learn a lot.
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kevincan
I'm the GMAT trainer at Kaplan, Madrid, Spain and I make these questions up for my students. (They like them). Do people mind if they are not OG or GMATPrep questions? I know they aren't easy, but I really like the thinking/explanations and I think people can learn a lot.


awesome.. GMAT trainer..
These are really quality questions, and should be helpful to everyone. Keep up the goood work.



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A list contains the surnames of six people, listed in order of age from oldest to youngest. If no two people have the same surname or the same birthday, what is the approximate probability that the names could be made to be in alphabetical order by switching exactly one name with another?

(A) 0.01 (B) 0.02 (C) 0.03 (D) 0.05 (E) 0.08



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Not pretty sure about this question
Total arrangements of names in alphabetical order is 6!=720. Selecting 2 out of 6 in 15 ways. The required prob is 15/720 or approx 0,02



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