Question 847186: how many combinations using $1, $.50, $.20, $.10 to make $2.60
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! The answer is 558. Spoiler...there is a lot of work shown below. Please let me know if it becomes too confusing to follow. We first break the $1 into 3 cases, then from there we break the $.50 into 12 cases, then we break the $.20 from there into 70 cases, and then the $.10 is used to form 558 distinct cases. This could be achieved using a tree diagram.
Let's first find all possible numbers of each dollar that can be used to construct $2.60.
$1: We can have 0,1,or 2 dollars without exceeding $2.60
If 0 $1, then we have $2.60 that we must create without using dollars.
If 1 $1, then we have $1.60 that we must create without using dollars.
If 2 $1, then we have $.60 that we must create without using dollars.
We break into 3 subcases.
From $2.60 we can use 0,1,2,3,4,or 5 $.50
From $1.60 we can use 0,1,2,or 3 $.50
From $.60 we can use 0 or 1 $.50
Now we've broken into 12 subcases.
From $2.60 we can use 0 - 13 $.20 (14 subcases)
From $2.10 we can use 0 - 10 $.20 (11)
From $1.60 we can use 0 - 8 $.20 (9)
From $1.10 we can use 0 - 5 $.20 (6)
From $.60 we can use 0 - 3 $.20 (4)
From $.10 we can use 0 $.20 (1)
From $1.60 we can use 0 - 8 $.20 (9)
From $1.10 we can use 0 - 5 $.20 (6)
From $.60 we can use 0 - 3 $.20 (4)
From $.10 we can use 0 $.20 (1)
From $.60 we can use 0 - 3 $.20 (4)
From $.10 we can use 0 $.20 (1)
Now we have 70 subcases (don't worry this is the last time)
From $2.60 we have 26 subcases
From $2.40 we have 24 subcases
From $2.20 we have 22
From $2.00 we have 20
From $1.80 we have 18
From $1.60 we have 16
From $1.40 we have 14
From $1.20 we have 12
From $1.00 we have 10
From $0.80 we have 8
From $0.60 we have 6
From $0.40 we have 4
From $0.20 we have 2
From $0.00 we have 0
From $2.10 we have 21
19
17
15
13
11
9
7
5
3
1
16
14
12
10
8
6
4
2
0
11
9
7
5
3
1
6
4
2
0
1
16
14
12
10
8
6
4
2
0
11
9
7
5
3
1
6
4
2
0
1
6
4
2
0
1
Adding this all together gives us 558.
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