You obviously need at least 2 pennies so this problem becomes 13 coins totaling 90 cents.

Lets say there are no half dollars in the coins.

You can't have more than 3 quarters (.25 *4 > .90) so the problem of 13 coins in 90 cents becomes the sum of 4 sub problems:

a problem for 0, 1, 2, or 3 quarters.

You can then repeat this process for the next highest denomination. This both eliminates 1 type of coin and reduces the amount to make each iteration.

The sub problems would be:

Using only pennies, nickels, and dimes make:

$.90 using 13 coins

$.65 using 12 coins

$.40 using 11 coins

$.15 using 10 coins

Doing some quick arithmetic you can quickly realize the last problem ($.15 using 10 coins) has no solutions so we can skip that one. Repeating the process for the 2nd to last one gives use cases for 0-4 dimes and so on.

Trying out we find the following solutions:

2 pennies, 5 dimes, 8 nickels

7 pennies, 1 quarter, 5 dimes, 2 nickels

2 pennies, 1 quarter, 1 dime, 9 nickels

The half dollar adds a couple more cases:

12 pennies, 1 half dollar, 1 quarter, 1 nickel

7 pennies, 1 half dollar, 7 nickels