Combinatorics problem - randomly distributing chewing gums In a kindergarden we distribute randomly $60$ chewing gums.
$20$ of those gums are red, $20$ are blue and $20$ are green.
Each kid receives exactly $2$ chewing gums.
Let $X$ be the number of kids who did not receive a red chewing gum, calculate $P\{X=10\}$.
Basically what I was thinking:
Denominator: $\dfrac{60!}{2^{30}}$.
Reasoning: We start with $60$ chewing gums, we then choose $2$ for the first kid, then we have $58$ left, out of which we need to choose $2$, and so on.
Numerator: $\dfrac{40!}{2^{10}}\cdot \dfrac{40!}{2^{20}}$.
Reasoning: We have $40$ chewing gums to distribute, similarly to the distributing process we did for the denominator, we distribute chewing gums for $10$ kids. Then we repeat the process again, we distribute $40$ chewing gums to  the remaining $20$ kids.
The result makes no sense.
Would appreciate an assistance trying to figure out what I did wrong.
 A: This is a tricky problem.  First of all, if you were only looking at one child, the probability that that child did not receive any red chewing gum is
$$\frac{\binom{40}{2}}{\binom{60}{2}}. \tag1 $$
In (1) above, the denominator represents the total number of ways of selecting which two pieces of chewing gum could be presented to one child, while the numerator represents the total number of ways of selecting two pieces of non-red chewing gum for the child.

To attack your problem, I recommend a Combinatorics approach, which is (in effect) what you attempted.  I will compute
$$\frac{N\text{(umerator)}}{D\text{(enominator)}}, \tag2 $$
where in (2) above, $D$ represents the total number of ways that the $60$ pieces of chewing gum could be presented to the children, and $N$ represents the total number of ways of doing this, where exactly $(10)$ children do not receive red chewing gum.
For convenience, I will compute
$$D = \binom{60}{2} \times \binom{58}{2} \times \cdots \times \binom{4}{2} \times \binom{2}{2} = \frac{(60)!}{2^{(30)}}. \tag3 $$
This matches your computation.  Note that in (2) above, the numerator and denominator must be computed in a consistent manner.  For convenience, in (3) above, I regard the order that the children receive the gum as relevant.
That is, I distinguish between child-1 getting two pieces of red gum, child-2 getting two pieces of blue gum, and vice-versa.  Therefore, when I compute $N$, I must do so in a consistent manner.
Edit
My computation of $D$ also presumes that for example one piece of red chewing gum is distinguishable from another.  This presumption is (also) needed to justify the computation in (3) above.
There are $~\displaystyle \binom{30}{10}~$ ways of selecting exactly which children will not receive any red chewing gum.
Assume that the children who did not receive red chewing gum are child-1, child-2, ..., child-10.
The number of ways that this can occur is
$$\binom{40}{2} \times \binom{38}{2} \times \cdots \times \binom{24}{2} \times \binom{22}{2} = \frac{(40)!}{2^{(10)} \times [(20)!]}. \tag4 $$
Under the above hypothetical, child-11, child-12, ..., child-30, have not yet received any gum.  Also, at this point, there are $40$ pieces of gum left, of which exactly $(20)$ are red.
Edit
A trap here, is that for this hypothetical to work, each of the $(20)$ remaining children must each receive at least one piece of red chewing gum.  Otherwise, the number of children who did not receive red chewing gum, in this hypothetical would not be exactly equal to $(10)$.
So, you have $(20)$ children left, $(20)$ pieces of red chewing gum left, and $(20)$ pieces of non-red chewing gum left.  The number of ways that these $(40)$ pieces of chewing gum can be distributed so that each of these $(20)$ children receives $(1)$ red piece of gum and $(1)$ non-red piece of gum is
$$[(20)!] \times [(20)!]. \tag5 $$
That is, you imagine that the child-11, child-12, ..., child-30 are lined up, and that the red gum and non-red gum will also be lined up to hand to each child.

Final computation.
$$N = \binom{30}{10} \times \frac{(40)!}{2^{(10)} \times [(20)!]} \times [(20)!]^2$$
$$ = \frac{(30)!}{(10)! \times [(20)!]} \times \frac{(40)!}{2^{(10)} \times [(20)!]} \times [(20)!]^2 $$
$$ = \frac{[(30)!] \times [(40)!]}{[(10)!] \times 2^{(10)}}.$$
$$D = \frac{(60)!}{2^{(30)}}. $$
Therefore,
$$\frac{N}{D} = \frac{[(30)!] \times [(40)!]}{[(10)!] \times 2^{(10)}} \times \frac{2^{(30)}}{(60)!}$$
$$ = \frac{[(30)!] \times [(40)!] \times 2^{(20)}}{[(10)!] \times [(60)!]}.$$
A: In your modelling process, it appears that all candies are distinguishable; and the order in which a kid gets two candies is unimportant. This means your denominator is correct. However your numerator is incorrect. I think there are a couple of issues here: First, why are you distributing $40$ candies twice when there are only $60$ total candies. Second, the reasoning for the denominator does not work for the numerator since candies are only partially distributed in each step (i.e., in distributing just to the $20$ children who get a red candy, you haven't yet distributed all the candies, which means that $\frac{40!}{2^{10}}$ counts incorrectly). I think there may be other issues as well, but I am not entirely sure of the reasoning for the numerator.
To get a correct numerator for your model:
Pick $20$ kids who will each get one red candy. $30\choose {20}$.
Choose $20$ remaining candies to give to the "red" kids. $40\choose 20$.
Put the red candies in order. $20!$ (in your model all candies are distinguishable).
Put the other $20$ candies given to the "red" kids in order. $20!$.
Distribute the remaining $20$ candies to the remaining $10$ kids as you did for the denominator: $\frac{20!}{2^{10}}$
These should give you the pieces for a correct numerator. (The final answer should agree with the answer given by user2661923.)

Another solution would be given as follows: In this solution, candies of a given color are indistinguishable.  Each child will get one candy in each of her or his hands.
Grab a red candy, and put it randomly in a hand.
Grab another red candy, and put it randomly in a remaining hand.
Grab another red candy, and put it randomly in a remaining hand.
Do this for all $20$ red candies. The probability that no kid has gotten two red candies is given by:
$$\frac{60}{60}\cdot\frac{58}{59}\cdot\frac{56}{58}\cdot\frac{54}{57}\cdot\cdots\cdot\frac{22}{41}$$
This also agrees with the answer of user2661923.
