If you choose three jokes randomly from an inventory of 12 each month, what is the probability that, in any given month, at least one of the three jokes will be different from the jokes you told the month before? .
My Solution is as follows :
$N(S)$ = $\frac{12!}{3!9!}$ =$220$ which represents the different combinations of 12 jokes taken three at a time .
Now let $A$ be the event that the three jokes are diffrent then $N(A)= \frac{9!}{3!6!}=84$ .
Let $B$ be the event that the 2 jokes are diffrent then $N(B)= \frac{10!}{1!9!}=10$
Let $C$ be the event that the 1 joke is diffrent then $N(C)= \frac{11!}{2!9!}=55$
Then $P(A\cup B\cup C)=\frac{N(A\cup B\cup C)}{N(S)}=\frac{84+10+55}{220}=\frac{149}{220}=0.677$
Can someone tell me what is wrong with this solution ?
You are close. Your calculation of $N(A)$ is correct. For $N(B)$, note that the $2$ different jokes must come from the remaining $9$, which gives $\binom92$ possibilities, and the $1$ joke that is the same must come from the original $3$, which gives $\binom31$ possibilities. So in total, we have $$N(B) = \binom92\binom31 = \frac{9!}{2!\ 7!}\frac{3!}{1!\ 2!} = \frac{9\cdot8}{2\cdot1}\cdot\frac31 = 108.$$ Similarly, $$N(C) = \binom91\binom32 = 9\cdot3 = 27.$$ Therefore the numerator in $P(A\cup B\cup C)$ should be $$N(A)+N(B)+N(C)=84+108+27=219.$$
Now, as @AndreNicholas pointed out in his comment, a much faster way is to notice that the only way that there would not be at least one new joke is if all the jokes are the same as last time; the chance of this is $\frac1{220}$, so the solution is $1 - \frac1{220} = \frac{219}{220}$.