Question

I have some questions regarding combinations in finite. Here is my first one I am stumped on:

Answer 1

We will get the number of possible selections, and then subtract the number less than 25 cents.

We can choose the number of dimes 5 ways 0,1,2,3 or 4.
We can choose the number of nickels 4 ways 0,1,2 or 3.
We can choose the number of quarters 3 ways 0,1, or 2.

That's 5*4*3  = 60 selections 

Now we must subtract from the 60 the number of selections of coins that are less than 25 cents. These will involve only dimes and nickels. 

To get a selection of coin worth less than 25 cents:
If we use no dimes, we can use 0,1,2  on all 3 nickels.
That's 4 selections less than 25 cents. (that includes the choice of No coins at all in the 60, which we must subtract).

If we use exactly 1 dime , we can use 0,1,2, or all 3 nickels.
That's the 3 combinations less than 25 cents.

And there is 1 other selection less than 25 cents, 2 dimes and no nickels. 

So that's 4+3+1 = 8 selections which we must subtract from the 60.
 
Answer 60-8 = 52 selections of coins worth 25 cents or more.