A shelf has $10$ distinct books of which $3$ are Math,$ 4$ are English and $3 $ are Science books. In how many ways can you arrange these books such that all English books will be placed in the center?
I have tried to answer the arrangement where each kind of books will be together-> $(3!.4!.3!.3!=5,184)$ but this time I can not determine which factorial to include.
In total, there are $10!$ ways to arrange 10 books. But lets fix 4 English books in the centre. Now there are $6!$ ways to arrange the remaining items. How many ways can we arrange the 4 English books in the centre? That would be $4!$. So the answer would be $4!*6! = 17280.$
I will try to answer your question a different way. $n$ distinct objects can be arranged in $n!$ ways. $10$ distinct objects can be arranged in $10!$ ways. Let $E_1 E_2 E_3 E_4$ denote the four English books in the centre. The likelihood that $E_1$ is English is $\frac{4}{10}$; the likelihood that $E_2$ is English given that $E_1$ is English is $\frac{3}{9}$; the likelihood that $E_3$ is English given that first two are is $\frac{2}{8}$; and the likelihood that $E_4$ is English given that the first three are English is $\frac{1}{7}$.
Thus the final answer is $10! * (\frac{4}{10}*\frac{3}{9}*\frac{2}{8}*\frac{1}{7})=17280$.
