<s>[INST] How many 10-bit strings with more 0’s than 1’s? I have to pick the answer from:
a.512
b.386
c.256
d.252
e.none of these
The number of bit strings of length 10 with n 0's (or n 1's in fact):
is C(10,n) , where C(a,b) = a! / [(a-b)!b!] is the combinitorial function.
So the first answer is C(10,3) = 120
In the second, there could be 6,7,8,9, or 10 zeros, so:
C(10,6) + C(10,7) + C(10,8) + C(10,9) + C(10,10)
= 210 + 120 + 45 + 10 + 1 = 386
In the third, C(10,7) + C(10,8) + C(10,9) + C(10,10)
120 + 45 + 10 + 1 = 176
In the fourth,
C(10,3) + C(10,4) + C(10,5) + C(10,6) + C(10,7) + C(10,8) + C(10,9) + C(10,10)
= 2^10 - C(10,2) - C(10,1) - C(10,0)
= 1024 - 45 - 10 - 1
= 968 
So should i go with e. none of the above?
Thank you. 
 [/INST] Exactly half of the strings that don't have $5$ of each. How many have $5$ of each? $\binom{10}{5}=252$ how many are there in total? $2^{10}=1024$.
Therefore the answer is $\frac{1024-252}{2}=386$
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