Permutation and Combination

Hello all, would very much appreciate some help with this question:

A code consists of blocks of 10 digits, 4 are zeroes and 6 are ones. Calculate the number of such blocks in which the first and last digits are the same as each other.

My final working is (8!/(6! x 2!)) + (8!/(4! x 4!)) = 98. I had two cases, the first case was when it started and ended with "0" and the second case where it started and ended with "1". The working is basically the arrangement of the remaining 8 digits divided by the nondistinct numbers. Could someone please tell me if I did it the right way?

My second question is why is this working wrong?

(4C2)(8!/(6! x 2!))+(6C2)(8!/(4! x 4!))

What I can't get out of my head is why do we not need to include any the selection of the two "1"s and "0"s at the start and end. I was wondering if for example, if you were selecting the "1"s that will be at the start and at the end, because all 6 "1"s are identical, and thus there is only 1 choice?

Thank you!