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02 Jun 2020, 07:52
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
73% (01:47) correct
27%
(01:57)
wrong
based on 139
sessions
History
Date
Time
Result
Not Attempted Yet
What is the probability that when the letters of the word SERENDIPITY are randomly rearranged, the first alphabet of the resulting word is neither T nor E?
A. 10/11
B. 9/11
C. 8/11
D. 5/11
E. 4/11
A. 10/11
B. 9/11
C. 8/11
D. 5/11
E. 4/11
firas92
Current Student
Joined: 16 Jan 2019
Last visit: 02 Dec 2024
Posts: 616
Given Kudos: 142
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
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02 Jun 2020, 12:08
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Bunuel
Total Number of permutations of the word SERENDIPITY \(= \frac{11!}{2!2!}\)
Number of permutations with T as first alphabet \(= \frac{10!}{2!2!}\)
Number of permutations with E as first alphabet \(= \frac{10!}{2!}\)
So the number of possible permutations with neither T nor E being the first alphabet \(= \frac{11!}{2!2!}-\frac{10!}{2!2!}-\frac{10!}{2!}\) \(= \frac{10!}{2!}(\frac{11}{2}-\frac{1}{2}-1)\) \(= \frac{10!}{2!}(4)\)
Therefore, the probability that when the letters of the word SERENDIPITY are randomly rearranged, the first alphabet of the resulting word is neither T nor E \(= \frac{\frac{10!}{2!}(4)}{\frac{11!}{2!2!}}\)
\(= \frac{\frac{10!}{2!}(4)}{\frac{10!}{2!}(\frac{11}{2})} = \frac{8}{11}\)
Answer is (C)
03 Jun 2020, 14:14
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IMO C
Total # of words = 11!/ (2! * 2!)
P(first alphabet of the resulting word is neither T nor E) = 1 - P(first alphabet of the resulting word is T or E)
P(first alphabet of the resulting word is T) = (10!/2!*2!) / (11!/2!*2!) = 1/11
P(first alphabet of the resulting word is E) = (10!/2!) / (11!/2!*2!) = 2/11
P(first alphabet of the resulting word is neither T nor E) = 1 - (1/11 + 2/11) = 8/11
Total # of words = 11!/ (2! * 2!)
P(first alphabet of the resulting word is neither T nor E) = 1 - P(first alphabet of the resulting word is T or E)
P(first alphabet of the resulting word is T) = (10!/2!*2!) / (11!/2!*2!) = 1/11
P(first alphabet of the resulting word is E) = (10!/2!) / (11!/2!*2!) = 2/11
P(first alphabet of the resulting word is neither T nor E) = 1 - (1/11 + 2/11) = 8/11
