How many combinations of r natural numbers add up to n?

[mathworker]

Find the number of different combinations of $$r$$ natural.numbers that add upto $$n$$

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I tried this for quite a fair amount.of.time but.couldn't figure it out.(Punch)

 

[Plato2]

Re: a tough combonotrics problem

Find the number of different combinations of $$r$$ natural.numbers that add upto $$n$$

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I tried this for quite a fair amount.of.time but.couldn't figure it out.

This is known as the problem of Partitions of an Integer.
Here is a fair webpage on the topic.

If you want print material see Ivan Niven's Mathematics of Choice, chapter six.

 

[mathworker]

Thanks for the link,I have gone through it.
But as far as I understood partition function doesn't give the number of partitions of specific cardinality.I mean if we want only the partitions that contains $$r$$ terms for example or can we define a restricted partition function that can do the job?If we can define how can we approximate such restricted $$p(x)$$

 

[Plato2]

Thanks for the link,I have gone through it.
But as far as I understood partition function doesn't give the number of partitions of specific cardinality.I mean if we want only the partitions that contains $$r$$ terms for example or can we define a restricted partition function that can do the job?If we can define how can we approximate such restricted $$p(x)$$

Well I did say that the webpage is only fair. I dislike its notation.
I suggest that you try to find Niven's book.

Example: The number of partitions of 6 into 3 summands is three:
\begin{align*} 6 &= 1+1+4\\ &=1+2+3\\ &=2+2+2\end{align*}

That is p_3(6)-p_2(6).

There is a clear recursive definition of p_k(n).

 

[Evgeny.Makarov]

Find the number of different combinations of $$r$$ natural.numbers that add upto $$n$$

There is a page in StackExchange about this. Of course, the problem is tricky if "combination" is used in its technical sense to mean a set. In contrast, permutations (i.e., ordered sequences) of summands are called compositions (rather than partitions). Their number is simple to figure out.