The divisors arise because of the need to convert the number of ordered pairs into the number of unordered pairs. The difference is perhaps best illustrated by means of an example.
Consider having just 4 players: $A, B, C, D$ which are to be arranged into 2 pairs.
Using the logic of the quoted "solutions" there are $4C2 \times 2C2 = 6 \times 1 = 6$ ordered pairings. These 6 possibilities are
- $(AB,CD)$
- $(AC,BD)$
- $(AD,BC)$
- $(BC, AD)$
- $(BD, AC)$
- $(CD, AB)$
where $(x, y)$ means $x$ is the first pair and $y$ is the second pair.
However, if we ignore the order in which the pairs occur there are in reality only 3 possibilities since 1. and 6. both split the four players into identical pairings, as do 2. and 5., as do 3. and 4. Thus for 2 pairs of players there are 6 different pairings when pairings are ordered, but only 3 different pairings when the order in which the pairs occur is ignored.
More generally, when there are $n$ pairs of players, any specific set of $n$ pairs can be ordered in $n!$ ways ($n$ choices of which pair is first, $n-1$ for which pair is second, etc). Since this is true for any set of $n$ pairs it follows that for $n \times 2$ players the number of possible ordered pairs is equal to number of possible unordered pairs multiplied by $n!$.
The population on which the probability calculation (numerator divided by denoinator) is based is that of unordered pairs of players. However, the calculation of $12C2 \times 10C2 \times 8C2 \times 6C2 \times 4C2 \times 2C2$ (and the unstated calculation for black-black and white-white pairings of $6C2 \times 4C2 \times 2C2$) calculates the number of ordered pairs.
Since, the 12 players are divided into 6 pairs, the number of possible unordered pairs (denominator of the probability calculation) is calculated by dividing the number of ordered pairs by $6!$. Similarly, the numerator is concerned with number of possible white-white and black-black pairings so involves 6 players in 3 pairs and therefore $3!$ is used as a divisor.
Incidentally, the solution given should have squared the numerator so that the probability is
$$\frac{(\frac{6!}{2^3 3!})^2}{\frac{12!}{2^6 6!}}$$
However, the numerical value of 5/231 is consistent with the corrected formula.
Problem 1 is standard Stars and Bars. You solved it correctly. The number of ways is $\binom{30+12-1}{12}$, or equivalently $\binom{30+12-1}{30-1}$.
For Problem 2, I think we are to assume that glazed is one of the $30$ kinds, and that all glazed doughnuts (from that store) are identical. Ditto for chocolate.
So if we want at least $3$ glazed and at least $4$ chocolate, put these in the bag. We need to get a further $5$ from the $30$ kinds. Solve as in the first problem. The number of ways is $\binom{30+5-1}{5}$. Note that your answer contained this number. The $\binom{7}{3}$ that you multiplied by should not be there.
For Problem 3, things are pretty much the same, except we must choose $5$ doughnuts from the $28$ kinds remaining, since we don't want any more glazed or chocolate.
Best Answer
In b) part of 2nd question. There is nothing mentioned about balanced teams. So you have to consider other cases also.