Number of sequences formed from $1, 1, 1, 2,3,4,5,6$ in which all three $1$s appear before the $6$

Question

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?

Answer 1

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.

Answer 2

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 $$\frac{1}{4}\binom{8}{3}5! = \frac{8!}{4!}$$

Answer 3

1. Create the initial sequence: $\_1\_1\_1\_6\_$.
2. Permute the remaining numbers. There are 4! ways to do this.
3. Note, that there are 5 places, where you can put the remaining 4 numbers (places can be selected multiple times). After the stars and bars method there are $\binom{5+4-1}{4}=\binom{8}{4}$ ways to select them.
4. Finally there are $$4!\binom{8}{4}$$ ways of selecting the required sequence.