Question

In a bag of marbles there are 4 blue marbles, 3 red marbles, 5 green marbles. You pick three marbles out of a bag one at a time without replacement. How many ways can you pick two red marbles first and one blue marble last?

Answer 1

Let's begin by listing out the information given to us:

blue marbles = 4

red marbles = 3

green marbles = 5

Total marbles = 4 + 3 + 5 = 12

The number of ways one can pick two red marbles first and one blue last is given by:

$\begin{gathered} Pe=^{12}P_3+^{11}P_2+^{10}P_4 \\ Pe=\frac{12!}{(12-3)!}+\frac{11!}{(11-2)!}+\frac{10!}{(10-4)!} \\ Pe=\frac{12!}{9!}\times\frac{11!}{9!}\times\frac{10!}{6!} \\ Pe=1320\times110\times5040=731,808,000 \\ Pe=731,808,000 \\ \\ \therefore The\text{ total number of ways of doing this is }731,808,000 \end{gathered}$