It is given on page #21 of U.G. book : Combinatorics - A guided Tour, by David Mazur, here an example detailing chances of getting three-of-a-kind in 5 card hands.
The problem statement is : From a standard 52-card deck, how many different five-card hands are possible? How likely is it that you will be dealt three-of-a-kind?
Solution approach for the second part is:
Encode each hand as a 5-list : A B C D E; where :
A is the denomination of the three-of-a-kind
$\implies$ 13 ways to chooseB is the 3-set of suits for the three-of-a-kind
$\implies$ $\binom{4}{3}$C is a 2-set of denominations for the other two cards
$\implies$ $\binom{12}{2}$D is the suit of the smaller denomination in C
$\implies$ $4$ ways to choose DE is the suit of the larger denomination in C
$\implies$ $4$ ways to choose E
The chances are given by the product rule.
My doubt is in question #24 following that. It asks to find wrong with the below alternate approach:
There are $52$ ways to select the first card that is a part of the three-of-a-kind. Then there are $\binom{3}{2}$ ways to pick the other two cards to make up the three-of-a-kind. Finally, there are $\binom{48}{2}$ ways to pick the other two cards (any cards except those of the denomination of the three-of-a-kind). By the product principle there are $52\cdot \binom{3}{2} \cdot \binom{48}{2}$.
My answer needs vetting, and is stated below in two parts. The first part concerns the first two steps:
(i) To have $52$ ways for the first card seems fine, & so is the next step to have $\binom{3}{2}$ ways to pick the other two cards.
But the issue is that the product of first steps (by the first approach) leads to: $13\cdot \binom{4}{3} = 13\cdot 4 = 52$, while the second approach's first two steps lead to : $52\cdot \binom{3}{2} = 52\cdot 3=156$.
In order to explain the multiplication factor of $3$, consider a smaller example of smaller number of denominations (let, $3$) & the same number of (i.e., $4$) suits; as it is the # of suits that matter in terms of picking three-of-a-kind.
$$ \left[
\begin{array}{c|c|c}
&A&B&C&D\\
1&1A&1B&1C&1D\\
2&2A&2B&2C&2D\\
3&3A&3B&3C&3D\\
\end{array}
\right] $$
The duplication of choices, by factor of $\binom{3}{2}=3$, is shown below for the above small case, by considering only the first denomination (i.e. $1$), with 1A1B1C, 1A1B1D, 1A1C1D, 1B1C1D occuring $3$ times each.
I have shown only $4$ cases for the first row (i.e, denomination = $1$), but it can be easily shown that there are a total of $12(=4*3)\cdot 3= 36$ cases, of which only $12$ are original, explaining the overcount's multiplier factor of $\binom{3}{2}=3$.
$$ \left[ \begin{array}{c|c|c|c} 1A&1A1B1C&1A1B1D&1A1C1D\\ 1B&1B1C1A&1B1A1D&1B1C1D\\ 1C&1C1A1B&1C1A1D&1C1B1D\\ 1D&1D1A1B&1D1A1C&1D1B1C\\ \end{array} \right] $$
(ii) Next comes the difference between the first approach's last 3 steps, & the second approach taking $\binom{48}{2}$.
The first approach leads to: $(\binom{12}{2}=66) \cdot (\binom{4}{1}=4) (\binom{4}{1}=4) = 66*16 = 1056$,
while the second approach leads to: $\binom{48}{2}=24*47= 1128$.
Second part analysis: The definition of three-of-a-kind considers the denomination of the rest two cards to be different, irrespective of their suits.
If take $\binom{48}{2}$, then also consider the cases where the third card is of the same denomination as second. This leads to an extra number of cases, that need be subtracted.
I am stating here subtraction, as the number of chances for second case are found by application of sum rule.
To show how it works, by considering the original example; have extra $1128-1056 = 72$ cases.
Only $12$ denominations are available for the 4th & 5th cards, but the two cards need have different denominations. So, either can be of higher denomination than the other.
Assuming, that denomination $1$ is taken up by the three-of-a-kind; then $12$ denominations are left. In each denomination there are still $3$ more suits (for the over-calculation to occur) leading to $12*3=36$ cases, which by ordering should lead to $36*2=72$ cases.
a) Fourth card denomination = $2$, fifth card denomination choices are = $3, 4, \cdots, 13$, a total of $12$ choices.
b) Similarly, the chances based on other denominations of fourth card are : $11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1$.
This sums up to : $12 \cdot 6 = 72$ cases.
For the first part, we can cut down the multiple counting by first converting the problem to a problem that consider order to matter and then remove it.
$$52 \cdot \binom32 \cdot \frac{2!}{3!}=52.$$
For the second part, to reduce those that are from the same denomination:
$$12 \cdot \binom43=72$$