. Javier needs to exchange a dollar bill for
coins. The cashier has 2 quarters, 10
dimes and 10 nickels. We are asked to
determine how many possible
combinations of coins Javier could get
back if he must get back at least one of
the quarters.
Let’s start by assuming he uses only 1
quarter.
D N
7 1
6 3
5 5 That’s 5 combinations.
4 7
3 9
Now, assume he’s going to get both
quarters.

11 Combination of Coins

Step-by-step explanation:

You've answered part of the question already.

I'll continue from where you stopped.

Let D represents Dime

Let N represents Nickels

The answer starts from here

Let’s start by assuming he uses only 1

quarter.

D N

7 1

6 3

5 5

4 7

3 9

That’s 5 combinations.

Notice that the sum of each row is a consecutive number.

i.e 8,9,10...

Now, let's assume he's going to get both quarters.

For Dime,he has 10/2 possible dime for a start.

For nickel, we'll start from the least possible; 0

While Dime reduces by 1, Nickel increases by 2

So, we have

D N

5 0

4 2

3 4

2 6

1 8

0 10

That's 6 combinations

Total Possible Combination = 5 + 6

Total Possible Combination = 11 Combinations

452,489 to the nearest thousand is 452,000