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?

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  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?

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