i have 8 pennies, 5 quarters, and 3 nickels in my wallet. if i take them out of my wallet one at a time without replacement, what is the probability that the eighth coin is a nickel, and the twelfth is a quarter?
There are \(\frac{16!}{3!5!8!}\)different arrangements of the 16 coins. If you fix the 8th and 12th spots as being occupied by by a nickel and a quarter, then you have only 14 coins left to arrange and there are \(\frac{14!}{2!4!8!}\)possibilities for the remaining coints. So the probability we are looking for is\(\frac{\frac{14!}{2!4!8!}}{\frac{16!}{3!5!8!}}\), which I think simplifies to \(\frac{1}{16}\).To see what is happening, let's look at a simpler problem:
Suppose you only have 1 nickel (N), two quarters(Q), and 3 pennies(P). These coins can be arranged in 60 different ways (\(\frac{6!}{3!2!1!}\)).
The following diagram show the sixty different arrangements; of those 12 of them have a quarter in the 3rd and a penny in the 5th spots. so the probability of drawing a quarter on the third and a penny on the 5th draw is \(\frac{12}{60}=\frac{1}{5}\).
