# 28 Crayons , 4 kinds, 4 kids and an specific requirement.

I'm pretty confuse how to solve the following problem,

28 crayons of which four are red are divided randomly between four toddlers: Bobby, Suzy, Tommy, and Andy. Each child gets seven crayons. If Bobby has one red crayon, what is the probability that Andy has the remaining three?

I know that they asked to me to find the probability of P(Andy has the $3$ red | Bobby has $1$)

And the total outcome is :

denominator => $\dfrac{28!}{7! 7! 7! 7!} = 4.7251 \times 10^{14}$

But I have no clue to continue this problem. If someone can help me I will be apreciatted. Thanks!

Book final answer said: $0.0263$

Suppose you distribute $1$ red and $6$ non-red crayons to Bobby. That leaves $3$ red and $18$ non-red, of which Andy gets $7$ in total. So the probability Andy gets $n$ red is $\dfrac{{3 \choose n}{18 \choose 7-n}}{{21 \choose 7}}$.

For $n=3$ this gives $\dfrac{3060}{ 116280}=\dfrac{1}{38}\approx 0.0263$.