Pocket Change

If $x, y, z \in \mathbb{N_{0}}$ respectively represent the number of nickels, dimes, & quarters, then we have $0.05x + 0.10y + 0.25z = 1 \Rightarrow x + 2y + 5z = 20$ . If $z = 0, 1, 2, 3, 4$ , then the corresponding ordered-pairs $(x,y)$ compute to:

$z = 0 \Rightarrow x + 2y = 20, (x,y) = (0,10); (2,9); (4,8); (6,7); (8,6); (10,5); (12,4); (14,3); (16,2); (18,1); (20,0)$ (11 ways),

$z=1 \Rightarrow x+2y = 15, (x,y) = (1,7); (3,6); (5,5); (7,4); (9,3); (11,2); (13,1); (15,0)$ (8 ways),

$z=2 \Rightarrow x+2y = 10, (x,y) = (0,5); (2,4); (4,3); (6,2); (8,1); (10,0)$ (6 ways),

$z=3 \Rightarrow x+2y = 5, (x,y) = (1,2); (3,1); (5,0)$ (3 ways),

$z=4 \Rightarrow x+2y = 0, (x,y) = (0,0)$ (1 way).

Hence, there are $11+8+6+3+1 = \boxed{29}$ total combinations to make $\$1$ out of just nickels, dimes, and quarters.