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I tried it by selecting 2 men out of 8 for bow side,and then arrange them in 2! ways.This can be done in$ \binom{8}{2}$*2! ways,and the stroke side can be crewed in 6 ways.So the required no. of ways are $\binom{8}{2}$*2!*6=336. But,my answer is not matching.It's correct answer is 5760.

Please help me in pointing out my mistake.

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See we don't want to select people we want to arrange people. Hence total ways are $^4P_2\times$$^4P_1\times$$^5P_5=5760$ assuming boat has 4 rowing on each side.

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You can select which two of the five who can row either side should row Bow in ${5\choose2}=10$ ways. That determines which four row Bow an which four row Stroke. Now you have $4!$ ways of arranging the Bow side and $4!$ ways of arranging the Stroke side. So total $10\cdot4!\cdot4!=5760$.

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Alternately, select seats for the three primadonnas first, which can be done in $4\cdot3\cdot 4$ ways. Afterwards the 5 remaining rowers can be seated in $5!$ ways, for a result of $$ 4\cdot 3\cdot 4\cdot 5!=5760 $$

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2 men sitting on a bow side can sit in 4!/2! ways, 1 man sitting on stroke side can site in 4 ways,remaining 5 can sit anywhere in 5! ways so 4!/2!*4*5!= 5760

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My answer is best as it is using basic principle Consider 4 each on both sides Then for 1st side we require 2 people so no. Of ways is 4(for 1st person)*3(for other one) Then for 2nd side we require 1 person so he has 4 choice Then for remaining 5 places we have 5*4*3*2*1 this (arrangement is of remaining people)

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