<s>[INST] Number of sequences formed from $1, 1, 1, 2,3,4,5,6$ in which all three $1$s appear before the $6$ I want to find number of sequences which contains $2,3,4,5,6$ exactly once and $1$ exactly three times. Also all three $1$s should be placed before $6$.
Example $1)1,1,1,2,3,4,5,6\\2)1,2,3,4,1,1,6,5$
I think that $6$ can not be at $1$st, $2$nd, and $3$rd place.
If we fix $6$ at fourth place then all three $1$'s are fixed at first three places and there are $4!$ such sequences possible.
How to calculate for $6$ be fixed at $5$th to $8$th place?
 [/INST] HINT: I will explain it for sixth place and leave the rest to you:
Since $6$ is on sixth place, there are five places left before it and three of them must be filled by $1$'s. For the rest of two, we can choose from $2,3,4,5$ with $\binom{4}{2}$ and arrange them in those five places with $\frac{5!}{3!}$ (since $1$'s are identical). For the places after sixth place, we can rearrange them with $2!$ so if $6$ is on sixth place, there are
$$\binom{4}{2}\cdot\frac{5!}{3!}\cdot2!$$
ways.
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