mpingo wrote:

Can anybody solve this question using permutation combination method, if it is possible to do so.

Bill purchases an item and receives no change. Before the purchase, he has only a five dollar bill, two 10 dollar bill, and a twenty dollar bill. How many distinct possibilities are there for the total amount of his purchase?

a.3

b.4

c.6

d.9

e.10

You can solve it using Combinatorics though I think the easiest method here would be to work on distinct amounts.

Just quickly check for multiples of 5 from 5 to 45 - You can make only multiples of 5 with 5, 10 and 20 and the maximum you can make is 5 + 10 + 10 + 20 = 45

So imagine 4 currency bills: 5 ,10, 10, 20

Now, can you make 10? Yes; 15? Yes; 20 Yes... till 45. You can make all these amounts and hence there are 9 distinct amounts.

The reason this method is the best is that you work on distinct amounts, not on distinct bills. So there is no worry of getting the same amount in two cases and not realising it.

A Combinatorics Method:

Considering the 3 distinct bills:

Amounts made from 1 bill: 3C1 ways

5, 10, 20

Amounts made from 2 bills: 3C2 (any two out of 5, 10, 20) + 1 (10 and 10)

Amounts made: 15, 25, 30, 20

Total 6 distinct amounts

Amounts made from 3 bills: 3C3 + 2 (two 10s and one other)

Amounts made: 35, 25, 40

Amounts made from all 4 bills: 5+10+10+20 = 45

Total 9 distinct amounts.

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