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6 persons a, b, c, d, e, and f are to be lined up in a row.

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6 persons a, b, c, d, e, and f are to be lined up in a row.

by Max@Math Revolution » Sun Mar 06, 2016 2:55 am
6 persons a, b, c, d, e, and f are to be lined up in a row. If a, b, and c adhere always together, how many possible cases are there?

A. 24
B. 36
C. 48
D. 72
E. 144


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by regor60 » Mon Mar 07, 2016 12:12 pm
Treat the group ABC as "X".

How many permutations of X D, E, and F = [spoiler]4![/spoiler]

Answer is A
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by Brent@GMATPrepNow » Mon Mar 07, 2016 12:30 pm
Max@Math Revolution wrote:6 persons a, b, c, d, e, and f are to be lined up in a row. If a, b, and c adhere always together, how many possible cases are there?

A. 24
B. 36
C. 48
D. 72
E. 144
This question is somewhat ambiguous since it isn't clear whether A, B and C must always line up in that order (i.e., ABC) or whether they can line other ways as well (i.e., ABC, ACB, BAC, BCA, CAB and CBA)

If they must line up as ABC, then regor60's response (4! = 24) is correct.
If they can line up in other ways, then the correct response is (4!)(3!), where 3! accounts for the various ways in which A, B and C can be ordered.

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by Max@Math Revolution » Mon Mar 07, 2016 3:44 pm
6 persons a, b, c, d, e, and f are to be lined up in a row. If a, b, and c adhere always together, how many possible cases are there?

A. 24
B. 36
C. 48
D. 72
E. 144


-> Consider a,b,c as one character from (abc)def. Then, it becomes 4!. Since an order can change, multiply 3!, which is (4!)(3!)=144. Thus, E is the answer.
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