A crew of an $8$ oar boat has to be chosen out of $11$ people, five of whom can row on stroke side only, four on the bow side only A crew of an $8$ oar boat has to be chosen out of $11$ people, five of whom can row on stroke side only, four on the bow side only, and the remaining two on the either side. How many different selections can be made? * * * As a crew of $8$ people is to be selected, $4$ must be on the stroke side and $4$ must be on the bow side. As $7$ people can row on stroke side and $6$ people can row on the bow side. So total selections are $\binom{7}{4}\times\binom{6}{4}=525$ but the answer given is $145$, I do not know where i have gone wrong.

Let's go case by case according to where the bimodal folk are assigned (if they are assigned). Notation: $(a,b)$ means that $a$ of the flexible rowers are on the left and $b$ are on the right.

Case I. we use $2$ of them.

Ia. $(2,0)$. Then we get $\binom 52 \times \binom 44=10$

Ib. $(0,2)$. then we get $\binom 54 \times \binom 42 = 30$

Ic. $(1,1)$. then we get $\binom 53\times \binom 43= 40$

Case II. we use $1$ of them.

IIa. $(1,0)$ We get $\binom 53\times \binom 44=10$

IIb. $(0,1)$ We get $\binom 54 \times \binom 43=20$

Note: there are two ways to choose the flexible man so case II yields $60$ combinations all in all.

Case III. we use $0$ of them. we get $\binom 54\times \binom 44=5$

Combining we get a total of $\fbox {145}$