Carla has 10 coins, each of which is a nickel, a dime, or a quarter, or a penny. The total value of her coins is less than $1.00. How many different combinations of coins might Carla have?
My computer says this many sums:
p n d q
1 = (7, 1, 1, 1) == 47
2 = (6, 2, 1, 1) == 51
3 = (5, 3, 1, 1) == 55
4 = (4, 4, 1, 1) == 59
5 = (3, 5, 1, 1) == 63
6 = (2, 6, 1, 1) == 67
7 = (1, 7, 1, 1) == 71
8 = (6, 1, 2, 1) == 56
9 = (5, 2, 2, 1) == 60
10 = (4, 3, 2, 1) == 64
11 = (3, 4, 2, 1) == 68
12 = (2, 5, 2, 1) == 72
13 = (1, 6, 2, 1) == 76
14 = (5, 1, 3, 1) == 65
15 = (4, 2, 3, 1) == 69
16 = (3, 3, 3, 1) == 73
17 = (2, 4, 3, 1) == 77
18 = (1, 5, 3, 1) == 81
19 = (4, 1, 4, 1) == 74
20 = (3, 2, 4, 1) == 78
21 = (2, 3, 4, 1) == 82
22 = (1, 4, 4, 1) == 86
23 = (3, 1, 5, 1) == 83
24 = (2, 2, 5, 1) == 87
25 = (1, 3, 5, 1) == 91
26 = (2, 1, 6, 1) == 92
27 = (1, 2, 6, 1) == 96
28 = (6, 1, 1, 2) == 71
29 = (5, 2, 1, 2) == 75
30 = (4, 3, 1, 2) == 79
31 = (3, 4, 1, 2) == 83
32 = (2, 5, 1, 2) == 87
33 = (1, 6, 1, 2) == 91
34 = (5, 1, 2, 2) == 80
35 = (4, 2, 2, 2) == 84
36 = (3, 3, 2, 2) == 88
37 = (2, 4, 2, 2) == 92
38 = (1, 5, 2, 2) == 96
39 = (4, 1, 3, 2) == 89
40 = (3, 2, 3, 2) == 93
41 = (2, 3, 3, 2) == 97
42 = (3, 1, 4, 2) == 98
43 = (5, 1, 1, 3) == 95
44 = (4, 2, 1, 3) == 99
