Two different families A and B are blessed with equal number of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family B is $$\frac{1}{12}$$, then the number of children in each family is:

Let the number of children in each family be $$n$$. Hence, the total number of children belonging to the two families together is $$2n$$.

Because no child can receive more than one ticket, distributing the three tickets is equivalent to choosing exactly three distinct children out of the total pool. Thus,

$$\text{Total ways} \;=\;\binom{2n}{3}.$$

The required event is that all the tickets go only to the children of family B. Since family B has $$n$$ children, the number of favourable selections equals the number of ways of choosing any three children out of these $$n$$, namely

$$\text{Favourable ways} \;=\;\binom{n}{3}.$$

By definition of probability,

$$P(\text{all three to B}) \;=\;\frac{\binom{n}{3}}{\binom{2n}{3}}.$$

We are told that this probability equals $$\dfrac{1}{12}$$. Therefore,

$$\frac{\binom{n}{3}}{\binom{2n}{3}} \;=\;\frac{1}{12}.$$

First we write the combinations in factorial form. Recalling the formula $$\binom{r}{k}=\dfrac{r!}{k!(r-k)!},$$ we get

$$\frac{\dfrac{n(n-1)(n-2)}{6}}{\dfrac{2n(2n-1)(2n-2)}{6}}=\frac{1}{12}.$$

The common factor $$6$$ cancels out, giving

$$\frac{n(n-1)(n-2)}{2n(2n-1)(2n-2)}=\frac{1}{12}.$$

We now cancel one factor of $$n$$ present in both numerator and denominator:

$$\frac{(n-1)(n-2)}{2(2n-1)(2n-2)}=\frac{1}{12}.$$

Cross-multiplying,

$$12(n-1)(n-2)=2(2n-1)(2n-2).$$

Dividing both sides by $$2$$ to simplify,

$$6(n-1)(n-2)=(2n-1)(2n-2).$$

Now we expand each side. First,

$$(n-1)(n-2)=n^{2}-3n+2,$$

so

$$6(n-1)(n-2)=6(n^{2}-3n+2)=6n^{2}-18n+12.$$

Next,

$$(2n-1)(2n-2)=4n^{2}-6n+2.$$

Equating the two expressions, we have

$$6n^{2}-18n+12=4n^{2}-6n+2.$$

Moving every term to the left side yields

$$6n^{2}-18n+12-4n^{2}+6n-2=0,$$

which simplifies to

$$2n^{2}-12n+10=0.$$

Dividing by $$2$$ throughout,

$$n^{2}-6n+5=0.$$

We solve the quadratic equation $$n^{2}-6n+5=0$$ using the quadratic-formula result $$n=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$ Here $$a=1,\,b=-6,\,c=5$$, so

$$n=\frac{6\pm\sqrt{(-6)^{2}-4\cdot1\cdot5}}{2}=\frac{6\pm\sqrt{36-20}}{2}=\frac{6\pm4}{2}.$$

This gives two possible roots:

$$n=\frac{6+4}{2}=5 \quad\text{or}\quad n=\frac{6-4}{2}=1.$$

However, $$n=1$$ is impossible because we must be able to hand all three tickets to children of family B, so family B must have at least three children. Consequently, the only viable solution is $$n=5$$.

Therefore, each family has $$5$$ children.

Hence, the correct answer is Option B.