Let p,d, and n represent the numbers of pennies, nickels, and dimes,
respectively, and value in cents = v. Then the equation is

v = 1p + 5n + 10d

We can choose the value of p any of 5 ways, that is, 0,1,2,3,4.

For every one of those 5 ways of choosing the value of p,

we can choose the value of n any of 2 ways, that is, 0 or 1.

So that 5*2 ways of choosing the number of pennies and nickels.

For every one of those 5*2 ways of choosing the values for p and n,

we can choose the value of d any of 2 ways, that is, 0 or 1.

So that 5*2*2 or 20 ways of choosing the number of coins.

That includes choosing no coins at all.  If you don't count 0 cents
as an amount of money, then you subtract 1 for the possibility and
the answer is 19.

Use all the coins and you have 19 cents.  You can always take away
enough of the coins to make any amount of money for 1 cent through
19 cents.

So the answer is 19, or if you include the possibility of taking
them all away, and having 0 cents, as an amount of money (no money,
which is how much money you have when you're broke!), then you could
say there are 20 values of money.  You'll have to ask your teacher
whether it's 19 or 20.

Edwin