Question 778981
<pre>
I will assume that you necessarily have at least one of each
kind of coin.  (For instance, 3 quarters is not allowed, nor 
is 15 nickels, nor 7 dimes and 1 nickel.)  

If you have 1 quarter, you can have from 1 up through 4 dimes,
and the rest in nickels.  That's 4 ways.

If you have 2 quarters, you can have either 1 or 2 dimes and
the rest in nickels.  That's 2 more ways.

So there are a total of 4+2 or 6 combinations of coins you 
can possibly have:

               Number of    Number of      Number of
Combination#   Quarters       dimes         nickels
    1             1             1             8
    2             1             2             6
    3             1             3             4
    4             1             4             2
    5             2             1             3
    6             2             2             1

You might impress your teacher by showing her or him that
you can also make a TREE diagram for that problem,
like this:  (See if you can figure it out) 


{{{drawing(400,2600/7,0,13,-5,9,line(2,0,5,3),line(2,0,5,-3),
line(9,6,5,3),line(9,4,5,3),line(9,2,5,3),line(9,0,5,3),
line(9,0,11,0),line(9,-2,5,-3),line(9,-4,5,-3),line(9,6,11,6),
line(9,4,11,4),line(9,2,11,2),line(9,-2,11,-2),line(9,-4,11,-4),
locate(11,0+.3,75cents),locate(11,2+.3,75cents),locate(11,4+.3,75cents),
locate(11,6+.3,75cents),locate(11,-2+.3,75cents),locate(11,-4+.3,75cents),
locate(10,0+.5,2),locate(10,2+.5,4),locate(10,4+.5,6),locate(10,6+.5,8),locate(10,-2+.5,3),locate(10,-4+.5,1), locate(7.3,4.15,2),locate(7.3,2.95,3),locate(7.3,1.75,4),
locate(7.3,5.35,1), locate(7.3,-1.8,1), locate(7.3,-3,2),
locate(3.2,1.95,1),locate(3.2,-.7,2),locate(3.2,9,"#Q"),
locate(7.3,9,"#D"), locate(10,9,"#N"), locate(3.2,8,"|"),
locate(7.3,8,"|"),locate(10,8,"|")

  )}}}

Edwin</pre>