Question

How many bit strings of length 14 contains the following types

(I) atmost three is?

(ii) an equal number of 0s and 1s

(iii) an odd number of 1s

Answer 1

I. Atmost three

$C(n,r)=\frac{n!}{(n-r)!r!}\\ n=14\\ r≤3\\ =C(14,0)+C(14,1)+C(14,2)+C(14,3)\\ =1+14+91+364=470$

II. An equal number of 0s and 1s. This means that there will be 7 1s and 7 0s.

$n=14\\ r=7\\ =C(14,7)\\ =3432$

III. An odd number of 1s. This means that 1s can be 1,3,5,7,9,11,13.

$=C(14,1)+C(14,3)+C(14,5)+C(14,7)+C(14,9)+C(14,11)+C(14,13)\\ =14+364+2002+3432+2002+364+14\\ =8192$