Suppose we have 10 slots, and we can fill them using number 1 six times, and number 0 four times.
Q1) How many different numbers can be produced?
Q2) Out of q1's numbers, how many do not start with 01?
Q3) Out of q1's numbers, how many do not start with 01 and do not end with 10?
Do we start at Q1 with something like $\frac{10!}{2!\cdot2!\cdot2!\cdot2!\cdot1!\cdot1!\cdot1!\cdot1!\cdot1!\cdot1!}$ and go from there? The fact that it's a binary choice with limits makes me struggle. Can anyone help?
For (1), just count how many ways you can select $6$ of $10$ places -- the remaining four are fixed automatically. To answer (2), find out how many begin with $01.$ That's equivalent to counting how many $5$ selections from $8,$ or equivalently, how many $3$ from $8.$ The last should now be easy. Simply count those that start with $01$ or end in $10.$