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# Eight students were to be seated along two rows such that four student

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Re: Eight students were to be seated along two rows such that four student [#permalink]
GMATinsight wrote:
Bunuel wrote:
Eight students were to be seated along two rows such that four students will be seated in each of the two rows called A and B. Two of the eight students definitely want to be seated in row A while one of them definitely wants to be seated in row B. In how many different ways can the eight students be seated?

(A) 5,760
(B) 5,960
(C) 6,500
(D) 6,760
(E) 7,160

Let students are
PQRSTUVW

Let, P and Q want to be in Row A
R wants to be in Row B

i.e. we need only two more students for Row A (other than P & Q) and those two should be selected from remaining 5 students (P Q and R have fixed rows already)

i.e. ways to select those two students for row A = $$^5C_2 = 10$$

Now arrangements of first row can be done in 4! ways
Now arrangements of Second row can be done in 4! ways

Total Outcomes = $$^5C_2*4!*4!= 10*24*24 = 5760$$

­Why are you not selecting students for row B? (why do you assume the remaining ones is ok) - something like 5C3 for row B

Could you help Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 95354
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Re: Eight students were to be seated along two rows such that four student [#permalink]
1
Kudos
miquelperezn wrote:
GMATinsight wrote:
Bunuel wrote:
Eight students were to be seated along two rows such that four students will be seated in each of the two rows called A and B. Two of the eight students definitely want to be seated in row A while one of them definitely wants to be seated in row B. In how many different ways can the eight students be seated?

(A) 5,760
(B) 5,960
(C) 6,500
(D) 6,760
(E) 7,160

Let students are
PQRSTUVW

Let, P and Q want to be in Row A
R wants to be in Row B

i.e. we need only two more students for Row A (other than P & Q) and those two should be selected from remaining 5 students (P Q and R have fixed rows already)

i.e. ways to select those two students for row A = $$^5C_2 = 10$$

Now arrangements of first row can be done in 4! ways
Now arrangements of Second row can be done in 4! ways

Total Outcomes = $$^5C_2*4!*4!= 10*24*24 = 5760$$