How many unique combinations with $7$ digits can be formed in which there are $2$ sets of repeating numbers with the size $2$?

Question

Example: $[1,1,2,2,3,4,5]$. How many unique combinations are there? I started off with $7!$ but there is going to be repeated values so I decided to try and narrow it down but I kept getting stuck.

Answer 1

There are $\binom{7}{2}$ ways to chose the two digits that will be duplicated, and then $\binom{5}{3}$ ways to choose the remaining three digits, so $\binom{7}{2}\binom{5}{3}$ combinations.

Answer 2

You just need to divide by $2$ for double-counting the sequences with the $1$'s swapped, and also divide by $2$ for double-counting the sequences with the $2$'s swapped. Hence, you get:

$$\frac{7!}{2 \cdot 2}$$

Answer 3

Assuming you don’t mean sequences, i.e. if we have {1,1,2,2,3,4,5}={1,2,1,3,2,5,4} then you have $\binom{10}{2}$ ways to choose the repeating digits. Then you can choose the other three digits in $\binom{8}{3}$ ways. Total =$\binom{10}{2}\binom{8}{3}=45\cdot 84 =3780$.
If, however the order matters, then multiply it by another $\frac{7!}{2!2!}$ to account for the repetitions. Hence the answer in this case is $1260\cdot 3780 = \binom{8}{3} \binom{10}{2} \frac{7!}{2!2!} $