Assume that you have 7 dimes and 5 quarters (all distinct), and you select 4 coins. In how many ways can the selection be made so that at least 3 coins are dimes?In how many ways can the selection be made so that at least 3 coins are dimes?

Answer 1

I see 2 cases
1. 3 coins are dimes, the 4th is a quarter
2. all 4 coins are dimes

number of selections for case 1 : C(7,3)*C(5,1)
= 35*5 = 175
number of selection for case 2: C(7,4) = 35

number of "at least 3 dimes" is 175+35 = 210

btw, did you mean to ask the same question twice?