# 15 Coin Collection Probability Problem

Cecil has a 15 coin collection. Four coins are quarters, seven coins are dimes, three are nickels and one is a penny. For each scenario, calculate the total possible outcomes if Cecil randomly selects five coins.

Cecil selects an equal number of nickels and dimes

So I have figured out most of the situations, but am unsure on how to go about doing this one. I know if it were saying probability of drawing a nickel or dime, I would just divide 10/15, 9/14, etc. But I have not come across a problem like this one where it is asking the probability of getting an equal amount of each. Could somebody show me how to do one like this?

• Just do it with two cases. First case #n=#d=1, second case #n=#d=2 (there is no third case, since you have only 4 quarters)
– ctst
Commented Oct 5, 2016 at 19:01
• How do you set that up though? Commented Oct 5, 2016 at 19:08

Since there are four quarters, seven dimes, and four other coins, the number of ways of selecting one of the four quarters, one of the seven dimes, and three of the other four coins is $$\binom{4}{1}\binom{7}{3}\binom{4}{3}$$ and the number of ways of selecting two of the four quarters, two of the seven dimes, and one of the other four coins is $$\binom{4}{2}\binom{7}{2}\binom{4}{1}$$ Since the two cases are disjoint, the total number of ways of selecting an equal number of quarters and dimes when five coins are selected from the four quarters, seven dimes, three nickels, and one penny is $$\binom{4}{1}\binom{7}{3}\binom{4}{3} + \binom{4}{2}\binom{7}{2}\binom{4}{1}$$ The total number of ways of selecting five of fifteen coins is $$\binom{15}{5}$$ Hence, the probability of selecting an equal number of quarters and dimes is $$\frac{\dbinom{4}{1}\dbinom{7}{3}\dbinom{4}{3} + \dbinom{4}{2}\dbinom{7}{2}\dbinom{4}{1}}{\dbinom{15}{5}}$$