Card Probability using inclusion exclusion You are dealt $13$ cards from a shuffled deck of $52$ cards. Compute the probability that you get all
four cards of at least one denomination (all Aces, or all Kings, or all Queens, . . . , or all Twos).
The answer to this is $\frac{{13\choose 1}{48 \choose 9}}{{52 \choose 13}} + \frac{{13\choose 2}{44 \choose 5}}{{52 \choose 13}} + \frac{{13\choose 3}{40\choose 1}}{{52 \choose 13}}$
I'm having a hard time understanding this solution. I think it depicts the probability of choosing all four of a first denomination plus all four of a first and second denomination plus all 4 of a first, second and third denomination, but that would mean that the third fraction should be smaller than the first two, but it isn't
Where am I going wrong? Perhaps this is the probability of the first four of a first domination plus four of a second plus four of  a third, but in that case I do not understand why we do ${48\choose 9}$, ${44\choose 5}$ and ${40\choose 1}$ because isn't that populating the rest of our selection of $13$ choices? And if that's the case then why are we populating it $3$ times?
 A: The true answer before some simplification is the following: $$\frac{ {13 \choose 1}\color{red}{{3 \choose 3}}{48 \choose 9}}{ 52 \choose 13} +\frac{ {13 \choose 2}\color{red}{{6 \choose 6}}{44 \choose 5}}{ 52 \choose 13} + \frac{ {13 \choose 3}\color{red}{{9 \choose 9}}{40 \choose 1}}{ 52 \choose 13}$$
Edit, this answer calculates some combinations twice, as John correctly pointed out below.
Breaking into cases:
Case $1$
We have $13$ ways to pick the first card, so we know which denomination is to be repeated, then we have $3$ ways to pick the next three (to complete the denomination). Then the remaining $9$ cards are picked from the $48$ left over  in the deck.
Cases $2$ and $3$ work very similarly as well, and there can't be a fourth case because that would exceed $13$ total cards picked. (We would need at least $16$ cards to be able to have a fourth case)
If you still have a question please feel free to comment below.
A: The probability of at least one quad is the probability of exactly one quad, plus the probability of exactly two quads, plus the probability of exactly three quads.
The last term gives the probability of exactly three quads:
$$P_3 = \frac{{13 \choose 3}{40 \choose 1}}{{52 \choose 13}}.$$
Choose the three denominations with the quads, and choose the $13$th card from what's left.
For exactly two quads, it should be
$$P_2 = \frac{{13 \choose 2}\left[{44 \choose 5} - {11 \choose 1}{40 \choose 1}\right]}{{52 \choose 13}}$$
We choose two denominations with quads, and choose five cards such that they don't produce a third quad (because we counted those above). We do this by looking at all combinations of five cards, and subtracting out the ones that result in a quad.
Similarly for $P_1$:
$$P_1 = \frac{{13 \choose 1}\left[{48 \choose 9} - {12 \choose 1}{44 \choose 5} + {12 \choose 2}{40 \choose 1}\right]}{{52 \choose 13}}$$
Choose the denomination with the quad, and the other nine cards. The first term in square brackets double counts multiple quad combinations. We subtract out the ones that result in one additional quad with the second term, but we "subtract out too much" because this term includes the double quads! So we add them back in with the third term. (This is inclusion/exclusion.)
