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29 Jul 2010, 20:09
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
31% (02:24) correct
69%
(02:05)
wrong
based on 361
sessions
History
Date
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Not Attempted Yet
Find the number of selections that can be made taking 4 letters from the word "ENTRANCE".
(A) 70
(B) 36
(C) 35
(D) 72
(E) 32
(A) 70
(B) 36
(C) 35
(D) 72
(E) 32
29 Jul 2010, 20:24
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dungtd
This problem is not the one you'll see on real GMAT. As combinatorics problems on GMAT are quite straightforward. But still it could be good for practice.
We have 8 letters from which 6 are unique.
Possible scenarios for 4 letter selection are:
A. All letters are different;
B. 2 N-s and other letters are different;
C. 2 E-s and other letters are different;
D. 2 N-s, 2 E-s.
Let's count them separately:
A. All letters are different, means that basically we are choosing 4 letters form 6 distinct letters: 6C4=15;
B. 2 N-s and other letters are different: 2C2(2 N-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;
C. 2 E-s and other letters are different: 2C2(2 E-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;;
D. 2 N-s, 2 E-s: 2C2*2C2=1.
15+10+10+1=36
Answer: B.
General Discussion
08 Sep 2014, 22:58
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dungtd
Yes, there is no zing in the question. It involves laborious counting of cases and hence isn't a GMAT type of question.
