There are 2 ways to choose each of of the 6 remaining bits,
so the answer to a) is 26 = 64

Since 101 has weight 2, the last 6 bits must contain 3 1's
There are "6 places choose 3" or 6C3 = 20 

c) Have weight 5 or start with 101?
Case 1:  Have weight 5.

From the 9 places to have digits, there are 5 places to 
have 1's.

That's "9 choose 5" = 9C5 = 126

Case 2:  Start with 101

That's problem a), or 64 

Case 3.  Have weight 5 and start with 101.

That's problem b), or 20

Let A = the set of 9-bit strings with weight 5.
Let B = the set of 9-bit strings that start with 101.

N(A or B) = N(A)+N(B)-N(A and B)

N(A or B) = 126+64-20 = 170 

Edwin 

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