Suppose we are dealt five cards from an ordinary 52-card deck. What is the probability that 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 $(13)(12) {4 \choose 3}{4 \choose 2} = 3744$. 


Suppose we are dealt five cards from an ordinary $52$-card deck.  What is the probability that we get no pairs?

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 
$$\binom{13}{5}4^5 = \frac{52 \cdot 48 \cdot 44 \cdot 44 \cdot 36}{5!}$$

Suppose we are dealt five cards from an ordinary $52$-card deck.  What is the probability that we get a full house?

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, 
$$\Pr(\text{full house}) = \frac{\dbinom{13}{1}\dbinom{4}{3}\dbinom{12}{1}\dbinom{4}{2}}{\dbinom{52}{5}}$$
