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We have to place 50 objects of 3 types in a row with 50 places. There are 26 objects of type $a$, 16 objects of type $b$ and 8 objects of type $c$. I am working on a problem where objects of type $b$ needs to be placed in the middle, as in between place no 15(inclusive) and 35(exclusive). How can I find out the total number of ways to do that?

My strategy was to place type $b$ objects first. This can be done in $\binom{(35-15 = 20)}{16}$ ways and then place remaining type $a$ and type $c$ objects in $\frac{34!}{26!\cdot8!}$. The total number becomes $\left(\binom{(35-15 = 20)}{16} \cdot \frac{34!}{26!\cdot8!}\right)$ ways. Is my reasoning correct?

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  • $\begingroup$ Your answer is correct. $\endgroup$ – N. F. Taussig May 18 at 9:20

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