Drawing numbers without replacement So if I were to draw 11 unique numbers from 0-99, (0 included), what is the probability that the last digit of all numbers drawn contain 0-9. (i.e a good outcome would be 10,11,12,13,14,15,16,17,18,19,20 as the last digits of 0-9 are selected).
Would I need to use the inclusion exclusion principle?
Is the answer to this question $1- \dfrac{\binom {10}{1} \binom {90}{11}- \binom {10}{2} \binom {80}{11}+ \binom {10}{3} \binom {70}{11}- \binom {10}{4} \binom {60}{11}+ \binom {10}{5} \binom {50}{11} - \binom {10}{6} \binom {40}{11} + \binom {10}{7} \binom {30}{11} - \binom {10}{8} \binom {20}{11} + \binom{10}{9} \binom{10}{11}}{\binom{100}{11}}$?

Your answer is correct, but there is really no need to use the devious route of inclusion-exclusion.
There are $10$ ways to choose the digit that will occur twice, 
and the rest is just application of the "normal" hypergeometric formula, thus
$10\times\left[\frac{\dbinom{10}2\dbinom{10}{1}^9}{\dbinom{100}{11}}\right]$
