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?

Here are some simpler approaches.
Method 1:  Place the $2$, $3$, $4$, and $5$ in $$\binom{8}{4}4! = \frac{8!}{4!}$$ ways.  Once you have done so, there is only one way to place the three $1$s and $6$ in the remaining positions so that all the $1$s appear before the $6$.
Method 2: Choose four positions for the three $1$s and the $6$, which can be done in $\binom{8}{4}$ ways.  There is only one way to arrange them in these positions so that all the $1$s appear before the $6$.  The remaining four distinct numbers may be placed in the remaining four positions in $4!$ ways.  Hence, there are 
$$\binom{8}{4}4! = \frac{8!}{4!}$$
admissible arrangements of the numbers.
Method 3: First, we count the number of distinguishable arrangements, then use symmetry to determine the number of admissible arrangements.  
Number of distinguishable arrangements:  Choose three of the eight positions for the $1$s, which can be done in $\binom{8}{3}$ ways.  The remaining five distinct numbers can be placed in the remaining five positions in $5!$ ways. Hence, the number of distinguishable arrangements of $1, 1, 1, 2, 3, 4, 5, 6$ is 
$$\binom{8}{3}5! = \frac{8!}{4!}$$
Number of admissible arrangements:  By symmetry, in one fourth of these arrangements will the $6$ appear after all three $1$s.  Hence, the number of admissible arrangements is 
$$\frac{1}{4}\binom{8}{3}5! = \frac{8!}{4!}$$
