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?
Need answer AND Explination as soon as possible please!!!!! Thank you!!!!!!
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