10d + 5n < 100
2d + n < 20
n < 20 - 2d
And since we must have at least 1 nickel,
0 < n < 20 - 2d
If there is 1 dime, d = 1, that inequality becoms
0 < n < 18 and there are 17 ways to have that.
If there are 2 dimes, d = 2, that inequality becoms
0 < n < 16 and there are 15 ways to have that.
If there are 3 dimes, d = 3, that inequality becoms
0 < n < 14 and there are 13 ways to have that.
And it goes down 2 each time we add a dime, and we
can have up to 9 dimes, so the answer is
17+15+13+11+9+7+6+5+3+2+1 = 81 ways.
Edwin
