Problem 428

Since $\$10 = 2\cdot \$5$, then $10 can be formed using 0, 1 or 2$5 bills and cannot be formed using more than 2 $5 bills.

Using 2 $5 bills, we obtain$10 exactly. There is no choice in this case, and so this gives exactly 1 way to make $10.

Using 1 $5 bill, we need an additional$5.

Since 5 is odd and any amount of dollars made up using 2 coins will be even, we need an odd number of 1 coins to make up the difference.

We can use 5 $1 coins and 0$2 coins, or 3 $1 coins and 1$2 coin, or 1 $1 coin and 2$2 coins to obtain $5.

This is 3 more ways.

Using 0 $5 bills, we need an additional$10.

Since 10 is even and any amount of dollars made up using 2 coins will be even, we need an even number of 1 coins to make up the difference.

The numbers of $1 coins and$2 coins that we can use are 10 and 0, 8 and 1, 6 and 2, 4 and 3, 2 and 4, or 0 and 5.

This is 6 more ways.

In total, there are $1+3+6=10$ ways in which André can make $10.