How many bit strings of length 5 contain at least two 1s?

We first find number of bit strings of length 5 that  contain less than two 1s.

(1) When there is no 1s in bit strings of length 5. (contain only zero)

Number of such string = ${}^{5}C_0 = \frac{5!}{0!(5-0)!} = \frac{5!}{5!} = 1$            $\left[{}^{n}C_r = \frac{n!}{r!(n-r)!}\right]$

(2) When there is only one 1s in the bit strings of length 5.

Number of such strings = ${}^{5}C_1 = \frac{5!}{1!(5-4)!} = 5$

Hence we have 6 strings ( from (1) and (2) ) that contain less than two 1s. The rest of the strings contain at least two 1s.

Also

the total number of bit strings of length 5  = $2^5 =32$