a + b + c = 10. a, b, c are whole numbers. Now this is similar to the previous question that we solved by placing 10 sticks and simplifying.
We cannot follow an exactly similar approach, as in this case a, b and c can be zero. Let us modify the approach a little bit. Let us see if we can remove the constraint that a, b, c can be zero.
If we give a minimum of 1 to a, b, c then the original approach can be used. And then we can finally remove 1 from each of a, b, c. So, let us distribute 13 sticks across a, b and c and finally remove one from each.
a + b + c = 13. Now, let us place ten sticks in a row
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This question now becomes the equivalent of placing two '+' symbols somewhere between these sticks. For instance
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This would be the equivalent of 4 + 5 + 4. or, a = 4, b = 5, c = 4.
There are 12 slots between the sticks, out of which one has to select 2 for placing the '+'s.
The number of ways of doing this is ^{12}C_{2}.
Correct Answer: ^{12}C_{2} = 66.
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