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19 May 2017, 01:53
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
66% (01:39) correct
34%
(01:48)
wrong
based on 560
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Result
Not Attempted Yet
5 people including A and B line up in a row. How many possible cases are there such that at least one person stands between A and B?
A. 24
B. 36
C. 48
D. 60
E. 72
A. 24
B. 36
C. 48
D. 60
E. 72
19 May 2017, 02:08
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One way of finding this is to find total ways of standing 5 people and then from those, subtracting those ways where A and B are together.
Total ways of 5 people standing in a row = 5! = 120
If A and B are together, treat them as 1 unit. Number of ways of arranging them in a row = 4! * 2! = 48
Thus, required number of ways = 120-48 = 72.
Hence answer is E
Total ways of 5 people standing in a row = 5! = 120
If A and B are together, treat them as 1 unit. Number of ways of arranging them in a row = 4! * 2! = 48
Thus, required number of ways = 120-48 = 72.
Hence answer is E
General Discussion
19 May 2017, 06:26
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MathRevolution
Here's a different approach:
Let's let the 5 people be A, B, C, D and E
Take the task of arranging the 5 people and break it into stages.
Stage 1: Arrange C, D and E in a row
We can arrange n items in n! ways.
So, we can arrange 3 people in 3! (aka 6) ways.
We can complete stage 1 in 6 ways
IMPORTANT: For each of the 6 arrangements of C, D and E, we can place a space on either side of each person.
For example, for the arrangement DEC, we can add spaces as follows: _D_E_C_
We can now place A in one of these spaces, and place B in one of these spaces.
This will ensure that A and B do NOT sit together
Stage 2: Place person A in one of the spaces.
There are 4 spaces to choose from. So, we can complete stage 2 in 4 ways
Stage 3: Place person B in one of the spaces.
There are 3 spaces remaining.So, we can complete stage 3 in 3 ways
By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus arrange all 5 people) in (6)(4)(3) ways (= 72 ways)
Answer:
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.
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