Denominations used: cent, 1¢=$0.01, nickel 5¢=$0.05, dime 10¢=$0.10, quarter 25¢=$0.25.
We can't get the sum of $1 without duplicating some of the coins, because there are only 4 regular denominations. Let c, n, d and q be the numbers of cents, nickels, dimes and quarters respectively.
c+n+d+q=15, so c=15-n-d-q.
c+5n+10d+25q=100 cents in value,
15-n-d-q+5n+10d+25q=100,
4n+9d+24q=85.
So q must be less than 4:
q=3: 4n+9d=13⇒n=d=1, c=10;
q=2: 4n+9d=37⇒d=1, n=7, c=5;
q=1: 4n+9d=61⇒d=5, n=4, c=5 and d=1, n=13, c=0;
q=0: 4n+9d=85⇒d=9, n=1, c=5 and d=5, n=10, c=0.
So we have:
No quarters, 9 dimes, 1 nickel, 5 cents=$0.90+$0.05+$0.05=$1.00;
No quarters, 5 dimes, 10 nickels, no cents=$0.50+$0.50=$1.00;
1 quarter, 5 dimes, 4 nickels, 5 cents=$0.25+$0.50+$0.20+$0.05=$1.00;
1 quarter, 1 dime, 13 nickels, no cents=$0.25+$0.10+$0.65=$1.00;
2 quarters, 1 dime, 7 nickels, 5 cents=$0.50+$0.10+$0.35+$0.05=$1.00;
3 quarters, 1 dimes, 1 nickel, 10 cents=$0.75+$0.10+$0.05+$0.10=$1.00.
If half-dollar is also included then:
h+q+d+n+c=15, c=15-h-q-d-n,
50h+25q+10d+5n+15-h-q-d-n=100,
49h+24q+9d+4n=85.
So h<2:
h=1: 24q+9d+4n=36⇒q=1, d=0, n=3, c=10 and q=0, d=4, n=0, c=10 and q=0, d=0, n=9, c=5;
We have:
1 half-dollar, 1 quarter, no dimes, 3 nickels, 10 cents=$0.50+$0.25+$0.15+$0.10=$1.00;
1 half-dollar, no quarters, 4 dimes, no nickels, 10 cents=$0.50+$0.40+$0.10=$1.00;
1 half-dollar, no quarters, no dimes, 9 nickels, 5 cents=$0.50+$0.45+$0.05=$1.00.
The half-dollar along with the other denominations gives us 9 ways of making up $1 with 15 coins.