How many bit strings of length 10 contains... I have a problem on my home work for applied discrete math
How many bit strings of length 10 contain
A) exactly 4 1s
the answer in the book is 210
I solve it 
$$C(10,4) = \frac{10!}{4!(10-4)!} = 210
$$
for the last 3 though I can't even get close. Did I even do part A right or is the answer only a coincidence?
B) at most 4 1s
the answer in the book is 386
C) at least 4 1s
the answer in the book is 848
D) an equal number of 0s and 1s
the answer in the book is 252

For part A it will be simply $C(10,4)=210$
For part B it will be $C(10,0)+C(10,1)+C(10,3)+C(10,4)=386$ because at most means maximum which we actually start from zero till that particular number
For part C it will be simply $C(10,4)+C(10,5)+C(10,6)+C(10,7)+C(10,8)+C(10,9)+C(10,10)=848$
For part D it will be simply $C(10,5)=252$
