You are given a deck of cards with 4 suits and 15 ranks, compute the probability of a "Straight", which is five cards in sequence, but not all from the same suit.
I have calculated that the number of possible hands is $ C(60, 5)$. Also the number of different rank I can get is $ C(11,1) $. However I am having trouble understanding the following question: "The suits can be anything other than all equal, so the number of possibilities for the suits of a straight is" If I was just calculating the number of possible straights I would have simply done $$4^5 * C(11,1)$$. Hence, I'd assume the question is asking for how many different suits combinations I can have. I tried $4^5$ and $4$. But both answers are wrong.
We have a deck of $4$ suits and $15$ different ranks for a total of $60$ cards. The hand is composed by $5$ cards.
If the hand is composed by consecutive ranks then it is a straight. I assume here that consecutive means that there is no cycle, by example the hand $14,15,1,2,3$ is not a straight.
Then, as the OP stated, the number of different straight by ranks is $11$, i.e. there are $11$ groups of cards, based on ranks, that are straight. From each group of straights $4$ combinations are prohibited, i.e. the straights with the same suit.
For each group of straights then there is a total of $4^5-4=1020$ valid straights. Then the probability to get a valid straight is
$$\frac{1020\cdot 11}{\binom{60}{5}}=\frac{935}{455126}\approx 0.2\%$$