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A bag contains 11 candy bars: three cost 50 cents each, four cost \$1 each and four cost \$2 each. How many ways can 3 candy bars be selected from the 11 candy bars so that the total cost is more than \$4?

Trinityvamp286  Jul 17, 2017
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Looking at each candy bar as a unique item, we have C(11,3) = 165  total ways to choose  any three of them

More than \$4

\$6 = Choose any three of the four \$2 ones  =  C(4,3) = 4 ways

\$5 = Choose any two of the four \$2 ones and any one of the four  \$1 ones  = C(4,2) * C(4,1)  = 24 ways

\$4.50 =  Choose any two of the four \$2 ones and any one of the three  \$.50 ones  = C(4,2) * C(3,1)  = 18 ways

So....the total ways are  4 + 24 + 18  =   46 ways

CPhill  Jul 17, 2017

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