.
The first thought that comes to a mind is (2+1)*(3+1) = 3*4 = 12 possible combinations,
providing 2+1 = 3 choices for pennies (1, 2, 0) and
3+1 = 4 choices for nickels (1, 2, 3, 0).
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| We ONLY should check that, combining this way, |
| we get all different values of the total sum. |
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For it, I created the corresponding rectangular table with two inputs.
One input is the most left vertical column for the number of pennies p = (0,1,2),
and the other input is the most upper line for the number of nickels n = (0,1,2,3).
The cells in the Table are the values of the total sum
Sum = 1*p + 5*n.
T A B L E
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0 1 2 3
0 0 5 10 15
1 1 6 11 16
2 2 7 12 17
As you see, the values in the Table are all UNIQUE, so we really have 3*4 = 12 opportunities,
if 0 (zero, ZERO) 0 pennies and 0 nickels is allowed;
or 12-1 = 11 possibilities, if this degenerated case is not allowed.
Solved.
