Find the number of hands of $13$ cards that contain:
(a) at least $1$ picture card (where a picture card has rank J, Q, K or A)
(b) $4$ cards of the same rank
(c) one card of each rank
For (a), I came up with answer as $\frac{16}{52} \times \frac{15}{51} \times…\times\frac{5}{40} $. Is there a quicker way to calculate this?
I also have in mind that I get to choose $13$ cards.. so there might be a different way to solve these kinds of questions?
Any help on how to tackle (b) and (c) will be appreciated!
I am a little confused as there are $13$ cards to draw but need to find the number of hands for one card of each rank.
Best Answer
For $(a)$, the number of such hands can be calculated as follows:$$X=\text{number of all possible hands}\\-\text{number of all possible hands with no picture card}\\=\binom{52}{13}-\binom{52-4\times 4}{13}$$
For $(b)$, we choose a 4-tuple of same-rank cards in $13$ different ways, leaving the other cards to be chosen independently; totally$$Y=13\times\binom{52-4}{13-4}$$
Edit
In case you want exactly 1 4-tuple, the answer is $$Y=13\times\binom{48}{9}-\binom{13}{2}\times\binom{44}{5}$$
For $(c)$, it is simply $$Z=4^{13}$$(why?)