Question

Compute the number of bit strings of length 14 that contain...
(a) exactly four 1s
(b) at most four 1s
(c) at least four 1s
(d) an equal number of 0s and 1s.

Answer 1

a) We should choose 4 place from 14 for 1. And another will fill with 0.

$\dbinom{14}{4}=\frac {14!} {4!(14-4)!}=1001$

b) We should choose 4, 3, 2, 1, 0 place from 14 for 1. And another will fill with 0.

$\dbinom{14}{0} + \dbinom{14}{1} + \dbinom{14}{2} + \dbinom{14}{3} + \dbinom{14}{4} =$

$1+14+91+364+1001=1471$

c) We have $2^{14}$ different bit strings of length 14. 1471 of them contains at most four 1 and 1001 contains four 1. Means bit strings of length 14 that contain at least four 1 is

$2^{14}-1471+1001=15 914$

d) We should choose 7 place from 14 for 1. And another will fill with 0.

$\dbinom{14}{7}=\frac {14!} {7!(14-7)!}=3432$