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03 Apr 2017, 00:40
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
74% (00:52) correct
26%
(01:26)
wrong
based on 137
sessions
History
Date
Time
Result
Not Attempted Yet
Two red bottles and three blue bottles are to be arranged in a straight line from left to right on a windowsill. How many different unique arrangements are possible?
A. 3
B. 5
C. 6
D. 10
E. 12
A. 3
B. 5
C. 6
D. 10
E. 12
divyamurali
Joined: 07 Nov 2016
Last visit: 06 Sep 2022
Posts: 6
Own Kudos:
Given Kudos: 33
Location: India
Schools: Great Lakes '19 (A)
GMAT 1: 620 Q47 V29
GPA: 3.9
03 Apr 2017, 01:09
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The answer is 10.
Total number of elements : 5 (2R, 3B)
Total number of combination: 5!
Taking redundant combination: 2! * 3!
therefore total number of unique combinations: 5!/(2!*3!)= 10
Total number of elements : 5 (2R, 3B)
Total number of combination: 5!
Taking redundant combination: 2! * 3!
therefore total number of unique combinations: 5!/(2!*3!)= 10
03 Apr 2017, 04:36
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Option D
5 Bottles = 2 Red bottles + 3 Blue bottles, can be arranged in a line in 5! ways.
Red bottles can be arranged among themselves in 2! ways
Blue bottles can be arranged among themselves in 3! ways
Hence, total no. of UNIQUE arrangements for 5 bottles = \(\frac{5!}{{2!*3!}} = 10\)
5 Bottles = 2 Red bottles + 3 Blue bottles, can be arranged in a line in 5! ways.
Red bottles can be arranged among themselves in 2! ways
Blue bottles can be arranged among themselves in 3! ways
Hence, total no. of UNIQUE arrangements for 5 bottles = \(\frac{5!}{{2!*3!}} = 10\)
