I would like to run this questions with stack:

How many bit strings of length 33 are there that start with 1010, end with 0101, and contain exactly 11 zeros.

  1. The amount of fixed bits are 8. Essentially we trying to choose $33-8=25$ bits.
  2. Must contain exactly 11 zeros, since there are already 4 zeros in the fixed end and beginning, we just need to choose 7 more zeros located in between the fixed lengths.

The number of ways arrange these bit strings adhearing to these rules are: ${25 \choose 7}$

  • 1
    $\begingroup$ Yup, you are correct. $\endgroup$ – user67773 Apr 17 '14 at 2:46

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