Suppose we are dealt five cards from an ordinary 52-card deck. What is the probability that
(c) we get no pairs (i.e., all five cards are different values)?
(d) we get a full house (i.e., three cards of a kind, plus a different pair)?
Solution:
(c) The number of ways this can happen is equal to $\frac{52 \cdot 48 \cdot44 \cdot 40 \cdot 36}{5!}$. Therefore $1317888/{52\choose5}$.
(d) The number of ways this can happen is equal to $(13)(12) {4 \choose 3}{4 \choose 2} = 3744$. Therefore the probability is $3744/{52\choose 5}$
I don't understand the solutions visually.
For (c) I don't understand why they do $52 \cdot 48 \cdot 44 \cdot 40 \cdot 36$. I suppose they divide by $5!$ because there are five categories of cards?
For (d) I don't understand how they got $(13)(12) {4 \choose 3}{4 \choose 2} = 3744$.
We must select cards from five of the thirteen ranks. For each selected rank, we must select one of the four suits. Hence, the number of favorable cases is $$\binom{13}{5}4^5$$ Since there are $\binom{52}{5}$ ways to select five of the $52$ cards in the deck, $$\Pr(\text{five cards of different ranks}) = \frac{\dbinom{13}{5}4^5}{\dbinom{52}{5}}$$
As for the given solution: The first card that is selected can be any of the $52$ cards in the deck. Since the second card that is selected must be of a different rank, it can be selected in $48$ ways. Since the third card that is selected must be of a different rank than each of the first two cards, it can be selected in $44$ ways. Continuing in this way, we get $52 \cdot 48 \cdot 44 \cdot 40 \cdot 36$ ordered selections of five cards of different ranks. However, the order of selection does not matter, so we must divide by the $5!$ orders in which the same five cards could be selected, so the number of favorable cases is $$\frac{52 \cdot 48 \cdot 44 \cdot 44 \cdot 36}{5!}$$ Dividing by $\binom{52}{5}$ gives the probability that each card is of a different rank.
You should check that $$\binom{13}{5}4^5 = \frac{52 \cdot 48 \cdot 44 \cdot 44 \cdot 36}{5!}$$
There are $13$ ways to select the rank from which three cards are selected and $\binom{4}{3}$ ways to select three of the four cards of that rank. There are $12$ ways to select the rank from which two cards are selected and $\binom{4}{2}$ ways to select two cards of that rank. Hence, the number of favorable cases is $$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}$$ Since there are $\binom{52}{5}$ ways to select five cards from the deck, $$\Pr(\text{full house}) = \frac{\dbinom{13}{1}\dbinom{4}{3}\dbinom{12}{1}\dbinom{4}{2}}{\dbinom{52}{5}}$$