How many permutations of the letters in HURRAH have the first R preceding the first H?
This is equivalent to the number of permutations with $R$ in the first position + The number of permutations where $R$ is in the second position, but $R,H$ is not first + The number of permutations where $R$ is third, but $R,H$ are not first or second
$\frac{5!}{2!}+2\frac{4!}{2!}+2!3!=96$ This answer doesn't seem that far fetched, is it right? What is a quicker answer?
The easiest approach here is probabilistic: you've got $\binom6{(2,2,1,1)}=180$ permutations in all, and the probability that the first letter from $\{H,R\}$ is $R$ is $\frac24=\frac12$. So the answer is $180*\frac12=90$. Note that this approach would also work if the letters $H,R$ had different frequencies (by adapting the probability of course).