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.
Best Answer
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 $\fbox {145}$