I'd like to calculate the probability of a domino player drawing 6 doubles out of 7 dominoes drawn.
My instincts wants to do the following to calculate:
There are 7 doubles and 21 nondoubles in a domino set of 28.
So...
$$ \frac{7}{28} \frac{6}{27} \frac{5}{26}\frac{4}{25}\frac{3}{24}\frac{2}{23}\frac{21}{22} $$
Which generalizes to $$ \frac{7!}{(7-k)!}\frac{21!}{(21-7+k)!}\frac{(28-7)!}{28!} $$ Where k is amount of doubles.
This incorrectly produces an answer that's a factor of $${7 \choose k}$$ from the - I believe - correct answer, as shown below:
There are 7 ways to have 6 doubles with an extra domino that can be any 21.
So...
$$ \frac{{7 \choose 6}*{21 \choose 1}}{{28 \choose 7}} $$
Which generalizes to
$$ \frac{{7 \choose k}*{21 \choose 7-k}}{{28 \choose 7}} $$
Or
$$ \frac{7!}{((k)!(7-k)!)} \frac{21!}{((21-7+k)!(7-k)!)} \frac{(28-7)!7!}{(28!)} $$
I'd like to understand intuitively why $${7 \choose k}$$ is needed? Is it simply because each domino is different? Would things be different if instead of 21 unique dominoes it's 21 of the same domino?
This is correct, except that it happens to express the exact probability of drawing the 6 doubles and one non-double, under the assumption that the non-double is drawn last.
The computation is off by a factor of $~7~$ because the non-double can be drawn in any of the $~7~$ draws.
So, the correct answer is
$$ 7 \times \frac{7}{28} \frac{6}{27} \frac{5}{26}\frac{4}{25}\frac{3}{24}\frac{2}{23}\frac{21}{22}. \tag1 $$
The more customary method of attack, that arrives at the same computation as that in (1) above, is to use combinatorics.
Express the answer as $~\dfrac{N}{D},~$ where $~N~$ represents the number of combinations of $~7~$ dominoes, which contain exactly $~6~$ doubles, and $~D~$ represent the total number of ways of selecting any $~7~$ dominoes, from the $~28~$ dominoes.
Note that in the previous paragraph, the computations of both $~N,~$ and $~D~$ assume that order of selection is irrelevant. Such an assumption is permitted in a combinatorics attack on a probability problem, as long as it is consistently applied (i.e. to both $~N~$ and $~D$).
Here, $~\displaystyle N = \binom{7}{6} \times \binom{21}{1},~$ because you have to determine which 6 doubles out of the 7 will be selected, and you have to determine which non-double, out of the 21 will be selected.
Then, $~\displaystyle D = \binom{28}{7},~$ which represents the number of different ways that 7 dominoes can be selected out of 28 dominoes.
Putting this all together, you have that the probability is
$$\frac{\binom{7}{6} \times \binom{21}{1}}{\binom{28}{7}}$$
$$= \frac{7!}{6! \times 1!} \times \frac{21!}{1! \times 20!} \times \frac{7! \times 21!}{28!}$$
$$= 7 \times 21 \times 7! \times \frac{21!}{28!}$$
$$= 7 \times 21 \times \frac{7!}{28 \times 27 \times \cdots \times 22}. \tag2 $$
The expression in (2) above is equivalent to the expression in (1) above.