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How many ways are there in which you can put in order $10 A's$,$6 B's $ and $5 C's$ without having two successive B's. When i say in order i mean something like that,example : AAAAAAAAAABBBBBCCCCCB.And what we dont want is to have any pairs of two consecutive B's (so we dont want to have BB ).For example the order of the letters AAAAAAAAAABBBBBCCCCCB is correct whereas the order AAAAAAAAAABBBBCCCCCBB is false because there are two consecutive B's: AAAAAAAAAABBBBCCCCC(BB).
This is a part of a longer exercise and for the most part i have used this formula:
$M(n_{1},....,n_{k})=\frac{n!}{n_{1}!...n_{k}!}$
$n=n_{1}+...+n_{k}$
so if for example i wanted the order of $10 A's$,$6 B's $ and $5 C's$ without any restrictions i would find it like this : $n_{1}=10$ because we have 10 A's,$n_{2}=6$ because we have 6 B's ,$n_{3}=5$ because we have 5 C's,so $n=10+6+5=21$ and the ways that exist are :$M(10,5,6)=\frac{21!}{10!6!5!}$

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It is more effective to realize that the space of configurations can be equipped with the structure of a cartesian product, so that the answer is given by the product of two binomial coefficients.

Remove the $B$s from a valid string: you get an anagram of $AAAAAAAAAACCCCC$. There are $\frac{15!}{10!5!}$ anagrams of such word. Now re-insert the $B$s back: we have to fit six $Bs$ in the sixteen spaces between consecutive characters, at the beginning or at the end of the previous string. The final outcome is so $$ \binom{15}{5}\binom{16}{6}=\color{red}{24048024}.$$

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