Using combinations to find how many 4-digit pins can equal to 20?

Question

I am unsure as to how to go about this problem. Any tips in helping me solve this would be greatly appreciated! I understand that I am to use combinations with repetition to solve such.

The question is asking: How many pins have digit sum 20? (a PIN is string a b c d of 4 decimal digits).

Using the combination with repetition formula I got 23C3 = 1771 ways (sorry I am new to stackexchange and don't know how to format)

Now I want to solve:

How many of the solutions to $a+b+c+d = 20$ have one of the 'digits' equal to $15$?

How many of the solutions to $a+b+c+d = 20$ have one or two of the 'digits' equal to $10$?

Thanks in advance!

How does a 'digit' equal 15 when we're only allowed to use decimal digits? — nickgard, Apr 23 '17 at 11:46
@PerudoJedi From my understanding, the question is trying to eliminate the cases where the values of a b c d are greater than 9. So by finding how many solutions where one of the digits equals 15, I can then eliminate it... ? — assassino, Apr 23 '17 at 12:16

nickgard · Accepted Answer · 2017-04-23T13:10:20.817

To answer your questions you can reuse the formula for combinations with repetition.

Any one of the four 'digits' can be $15$ and then the other three digits must sum to $5$, which gives $4\times\binom{7}{2}=84$ possible solutions.

Similarly, assume any one of the four 'digits' is $10$, then the other three digits must sum to $10$, so we have $4\times\binom{12}{2}=264$. However, this will double-count the cases where the four 'digits' are $10,10,0$ and $0$ in some order. There are $6$ of these so the answer is $264-6=258$ solutions.

Using combinations to find how many 4-digit pins can equal to 20?

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