Originally Posted by
somethingsmells 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!))