How many bit strings of length 14 have: (a) Exactly three 0s? (b) The same number of 0s as 1s? (d) At least three 1s?

(a) Exactly three 0s means other digits are always one, so the number of permutations would be equal to $\frac{14!}{(14-3)!}=14\cdot13\cdot12=2184$ . Answer is 2184.

(b) Since it's a bit string of 14 digits and there can be only digits 0 or 1, we need to select 7 0s and the rest would be 7 1s, so the number of permutations would be equal to $\frac{14!}{(14-7)!}=14\cdot13\cdot12\cdot...\cdot8=17297280$ . The answer is 17297280.

(c) The answer would be similar to question a, but here the rest of the digits can be either 0 or 1, so it would have to be multiplied by $2^{11}$ because there 11 digits and each one of them can be either of two choices. $2184\cdot2^{11}=4472832$ . Answer is 4472832.