Question

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?

Answer 1

(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.