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Five students, Adnan, Beth, Chao, Dan, and Edmund are to be

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Five students, Adnan, Beth, Chao, Dan, and Edmund are to be [#permalink] New post 30 Jul 2018, 11:05
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69% (01:12) correct 30% (00:50) wrong based on 13 sessions
Five students, Adnan, Beth, Chao, Dan, and Edmund are to be arranged in a line. How many such arrangements are possible if Beth is not allowed to stand next to Dan?

(A) 24
(B) 48
(C) 72
(D) 96
(E) 120
[Reveal] Spoiler: OA

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Re: Five students, Adnan, Beth, Chao, Dan, and Edmund are to be [#permalink] New post 31 Jul 2018, 07:03
If no restrictions are in place.
Total number of ways to arrange the five students is 5!
however to find the number of ways when 2 of the students are not next to each other;
we can find the arrangements of 5 people in which the 2 students could be seated together

i.e 4!*2!. here we assume beth and dan are fused together to form 1 person. Then we take out the number of ways in which the two of them can be arranged among themselves which is 2!

Hence 5! = 120
4!*2! = 48
Therefore our answer is 120 - 48 = 72
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GMAT Club Legend
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Joined: 07 Jun 2014
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Re: Five students, Adnan, Beth, Chao, Dan, and Edmund are to be [#permalink] New post 21 Aug 2018, 18:36
Expert's post
Expplanation

The number of ways in which the students can be arranged with Beth and Dan separated is equal to the total number of ways in which the students can be arranged minus the number of ways they can be arranged with Beth and Dan together.

The total number of ways to arrange 5 students in a line is 5! = 120. To compute the number of ways to arrange the 5 students such that Beth and Dan are together, group Beth and Dan as “one” person, since they must be lined up together. Then the problem becomes one of lining up 4 students, which gives 4! possibilities. However, remember that there are actually two options for the Beth and Dan arrangement: Beth first and then Dan or Dan first and then Beth. Therefore, there are (4!)(2) = (4)(3)(2)(1)(2) = 48 total ways in which the students can be lined up with Dan and Beth together.

Finally, there are 120 – 48 = 72 arrangements in which Beth is separated from Dan.

Alternatively, compute the number of ways to arrange the students directly by considering individual cases. In this case, investigate how many ways there are to arrange the students if Beth occupies each spot in line and sum them to find the total. If Beth is standing in the first spot in line, then there are 3 options for the second spot (since Dan cannot occupy this position), 3 options for the next spot, 2 options for the next spot, and finally 1 option for the last spot.

This yields (3)(3)(2)(1) = 18 total possibilities if Beth is first. If Beth is second, then there are 3 options for the first person (Dan cannot be this person), 2 options for the third person (Dan cannot be this person either), 2 options for the fourth person, and 1 option for the fifth. This yields (3)(2)(2)(1) = 12 possibilities. In fact, if Beth is third or fourth in line, the situation is the same as when Beth is second. Thus, there are 12 possible arrangements whether Beth is 2nd, 3rd, or 4th in line, yielding 36 total arrangements for these 3 cases.

Using similar logic, the situation in which Beth is last in line is exactly equal to the situation in which she is first in line. Thus, there are (18)(2) = 36 possibilities in which Beth is first or last. In total, this yields 36 + 36 = 72 possible outcomes when considering all of the possible placements for Beth.
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Re: Five students, Adnan, Beth, Chao, Dan, and Edmund are to be   [#permalink] 21 Aug 2018, 18:36
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