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I know there are some questions with solutions of the form

$\binom{n + k - 1}{k - 1}$

There are also questions with solution of the form

$\binom{n + k - 1}{k}$

and there are questions with solution of this form

$\binom{n + k}{k}$

I am trying to study for exam, and I am confused sometime with form of answer correspond to which type of question.

Could you please help me by giving typical examples which has the answer of each form?

Thank you a lot!

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These aren't always associated with cerain types. You may want to review coursebook materials which show you how they arrived at the binomial coefficients. A tiny wording difference in the question may yield different binomial coefficients. – Frenzy Li Oct 19 '12 at 2:29
up vote 1 down vote accepted

Mechanically matching formulas with problem types is a bad idea. In fact, it’s a good way not to acquire real understanding. And in purely practical terms, it doesn’t work very well, because any good instructor can design problems that use only the concepts that you’ve learned but that don’t look just like any of the problem prototypes that you’ve already seen.

I would say, in fact, that all three of the expressions that you list are essentially the same and are useful in solving the same kinds of problems. $\binom{n+k-1}{k-1}$ becomes $\binom{n+k}k$ when you substitute $k$ for $k-1$, and the latter becomes $\binom{n+k-1}k$ when you substitute $n-1$ for $n$. Here’s an example: $\binom{n+k-1}{k-1}$ is the number of solutions to the equation $x_1+x_2+\ldots+x_k=n$ in non-negative integers. $\binom{n+k}k$ is the number of solutions to the equation $x_0+x_1+\ldots+x_k=n$ in non-negative integers. But these are exactly the same kind of problem: the only difference is that in one problem I had $k$ unknowns numbered $1,\dots,k$, and in the other I had $k+1$ unknowns numbered $0,\dots,k$. Thus, it’s impossible usefully to associate any of these three binomial coefficients with a particular type of problem.

I’m afraid that in the long run (and to a very considerable extent in the short run as well) the only successful approach is to work to understand the reasoning behind the use of a particular expression or formula in solving a particular problem. Without this understanding, you’ll find your self blindly applying inapplicable formulas on the basis of superficial resemblances between problems; with it, you stand a reasonable chance of analyzing problems of the appropriate level of difficulty successfully.

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