Question

A person has ten bills in his pocket: 3 ones, 2 fives, 3 tens, and 2 twenties. If three bills are drawn at random without replacement then how many ways are there that their sum is $23 or more and with at least one "5" bill?

Answer 1

 

So, if I can’t use any $1 bills that means the standard formula of 3 x $20 and 3 x$1 isn’t available.

 

So, in the US what bills do we have below $100

$1

$2

$5

$10

$20

$50

SO, if I cant use singles then I have to come up with some combination that gets me to 63.

 

Using my extensive experience as a cashier, I am going to go with the $50 as I know none of the smaller bills are going to tally up to $63 without going over the 6 bill requirement.

 

So $50

 

Then let’s add a $10. That gets me to $60. But, I can’t use singles and I only have 2 bills used. So, there is no way to either get to $63 without using singles or to get to 6 bills in any case.

 

So, let’s add a $5 to make it $55. So, from there, I need 4 more bills and 8 more dollars. So, clearly, I need 4 x 2$ bills

 

$50+$5+$2+$2+$2+$2 = $63 and 6 bills.

$63