Question 132459
I don't have to know the exact amount of each combination, just
how many combinations there are, so I can ignore the pennies, and
just be aware that I'll need pennies to fill in the shortfalls 
when I have less that 66c
nickels
--------
(0-13)
I can have no nickels up to 13 nickels. 14 nickels would make 
{{{14*5 = 70}}}c which exceeds the change
How many dimes can I possibly have with each of (0-13) nickels?
and not exceed 66c?

nickels   dimes   subtotal
-------   -----   --------
0         (1-6)      6
1         (1-6)     12
2         (1-5)     17
3         (1-5)     22
4         (1-4)     26
5         (1-4)     30
6         (1-3)     33
7         (1-3)     36
8         (1-2)     38
9         (1-2)     40
10        (1)       41
11        (1)       42
12        (0)
13        (0)
---------------
To tally up what I have so far, 
(1) (0 nickels, 66 pennies)
(2) (1 nickel, 61 pennies)
. . . .
(14) (13 nickels, 1 penny)
(15) (0 nickels, 1 dime, 56 pennies)
. . . .
(56) (11 nickels, 1 dime, 1 penny)
---------------
Now I have to take into account the 
combinations that use quarters
---------------
nickels    dimes: (1)    (2)    (3)   (4)
-------    ------------------------------
0      quarters: (1-2)   (1)    (1)   (1)
1                (1-2)   (1)    (1)   (0)
2                (1-2)   (1)    (1) 
3                 (1)    (1)    (0)
4                 (1)    (1)
5                 (1)    (0)
6                 (1)
7                 (0)
This table adds 19 more possibilities, bringing the
subtotal to 75
---------------
Now for the 50c as part of the change:
---------------
nickels     dimes:  (0)   (1)
-------     -----------------
0             50c:  (1)   (1)
1                   (1)   (1)
2                   (1)   (0)
3                   (1)
4                   (0)
---------------
This table sadds 6 more combinations.
The final grand total is 81, unless I 
messed up somewhere.