# Basic probability homework question help (combinatorical counting problem)

I would really appreciate some help in the following probability question. I translated it so I'm sorry in advance for English mistakes.

There are $$6$$ children in a family and many candies in $$4$$ colors - blue, red, green, pink. The $$2$$ eldest children get to choose $$2$$ candies each. The $$4$$ other children get to choose $$1$$ candy apiece. (The same candy type can be chosen by a few children).

a. Count the number of choices.

b. What is the probability that exactly $$2$$ children choose blue candies?

My attempt at solution:

a. $$\binom{4}{2}^2 \cdot 4^4 = 6^2 \cdot 4^4$$

This solution is correct according to the solutions given to me.

b. I tried to solve it in the following way:

Dividing the question into 3 cases:

$$(1)$$ The number of cases in which the two eldest children choose blue candies: $$3^4$$

$$(2)$$ The number of cases in which two younger children choose blue candies: $$3^2 \cdot 3^2 = 3^4$$

$$(3)$$ The number of cases in which one blue candy is chosen by an older child and one by a younger: $$2 \cdot 4 \cdot 3 \cdot 3^3 = 2 \cdot 4 \cdot 3^4$$

I sum all the case to get $$10 \cdot 3^4$$

The final solution I get is: $$\frac{10 \cdot 3^4}{6^2 \cdot 4^4} = 0.088$$

My issues with this question is that in the solutions given to me, the number of options in case $$(1)$$ is calculated as: $$3^2 \cdot 3^4$$ and in case $$(2)$$: $$3^2 \cdot \binom{4}{2} \cdot 3^2$$

My question is - why is this solution correct?

Any help would be greatly appreciated.

$$4\choose 2$$