What is the solution of the recurrence? I have the following recurrence which I can not solve .
Trivial Case :
1. N(o,p) = 1 for p>=3
2. N(r,3) = 1 for r>=0
Recurrence :
1. N(r,p) = N(r,p-1) + N(r-1,p)

What is the value of N(r,p) ? I want to solve the recurrence and want to find the value of N(r,p) . Plz help me to find the value of this recurrence .     
 A: I think the comments really say what need to be said.  Nevertheless, I'd like to try to expound on this.
The question was to "find the value" of $N(r,p)$, but it has been granted that $N(r,p) = \binom{p+r-3}{p-3}$.  If the question is whether a better "solution" (or "value"?) exists, I would have to say no.  I would point you to a definition of Binomial Coefficients in case you are unfamiliar with the method of calculating these numbers.  
Perhaps the result $\binom{p+r-3}{p-3}$ does not satisfy because it does not resemble expressions such as $p+r$ or $\frac{p^2}{\sqrt{r}}$.  The fact is, we cannot express the binomial coefficients (in general) in terms of $+$, $-$, $\times$, $\div$, $\sqrt{}$, and/or exponents, unless we are willing to throw in some extra dots (ellipsis) to indicate a certain pattern:
$$
  \binom{n}{m} = \frac{n(n-1)(n-2) \cdots (n-m+1)}{m(m-1)(m-2)\cdots 1}
$$
Incidentally, your recurrence $N(r,p)$ could be expressed in this way by:
$$
  N(r,p) = \frac{(p+r-3)(p+r-4)\cdots (r+1)}{(p-3)(p-4) \cdots 1}
$$
To summarize, the best solution to your recurrence is $N(r,p) = \binom{p+r-3}{p-3}$.  It is what it is.
