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=(12!)/((12-3)!)+(11!)/((11-2)!)+(10!)/((10-4)!) \\ Pe=(12!)/(9!)*(11!)/(9!)*(10!)/(6!) \\ Pe=1320*110*5040=731,808,000 \\ Pe=731,808,000 \\ \\ \therefore The\text{ total number of ways of doing this is }731,808,000 \end{gathered}`