Bruce H. answered 02/01/23
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MATH/STAT Tutor Emphasizing Critical Thinking and Logical Reasoning
C(4,2)XC(3,1)+C(4,3)XC(3,0) = 6X3+4X1 = 18 + 4 = 22.
Ava A.
asked 01/30/23Assume that you have 4 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?
Bruce H. answered 02/01/23
MATH/STAT Tutor Emphasizing Critical Thinking and Logical Reasoning
C(4,2)XC(3,1)+C(4,3)XC(3,0) = 6X3+4X1 = 18 + 4 = 22.
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