Question

How many Combinations of bit strings length 9 have:                       a)    exactly three 0s? 

b)    at least seven 1s?

Answer 1

If set $A$ which contains $n$ elements consists of $n_1$ elements of the first kind, $n_2$ elements of the second kind, ..., and $n_k$ elements of $k$-th kind ($n=n_1+n_2+...+n_k$), the number of permutations with repetition is given by:

$\frac{n!}{n_1!\cdot n_2!\cdot ...\cdot n_k!}$

a) Let us find the number of bit strings length 9 that have exactly three 0s (and thus exactly six 1s):

$\frac{9!}{3!\cdot 6!}=\frac{9\cdot 8\cdot 7}{2\cdot 3}=84$

b) Let us find the number of bit strings length 9 that have  at least seven 1s:

$\frac{9!}{7!\cdot 2!}+\frac{9!}{8!\cdot 1!}+1=\frac{9\cdot 8}{2}+9+1=46$