Question

Assume that you have 7 dimes and 3 quarters (all distinct), and you select 3 coins. In how many ways can the selection be made so that at least 2 coins are dimes?

Answer 1

Well, I know that there is going to be a 70% chance to get a dime, but there is going to be a 3.5 chance out of 5 to get two dimes.

Answer 2

Which is also a 70%.

Answer 3

\[\binom{7}{2}\times \binom{3}{1}+\binom{7}{3}\]

Answer 4

first one is the number of ways to choose 2 out of the 7 dimes and 1 out of the 3 quarters
second is the number of ways to choose 3 out of the 7 dimes
do you know how to compute these? sometimes\(\binom{7}{2}\) is written as \(_7C_2\) or even \(^7C_2\)

Answer 5

Awesome, thank you so much. Yes, I do know how to compute them, just couldn't figure out if it was combination or permutation.

Answer 6

you can't tell the quarters apart in this problem, so definitely a combination