Number Theory and combinatorial Today, I took this observation from my note book. I am looking the strategy to deal this statement. The difference between $$\binom{n}{p}$$ and  $$\left\lfloor\frac{n}{p}\right\rfloor\,$$  is divisible by p for a positive integer n and p is prime with >1. Here $$\binom{n}{p}$$  is the number of ways one can choose p out of n elements and $$\left\lfloor{x}\right\rfloor\,$$  is the greatest integers not exceeding the real  number x.
The above one is I found from the following problem.
5 divides the difference between $$\binom{n}{5}$$and$$\left\lfloor\frac{n}{5}\right\rfloor\,$$
Numerically we can solve. I would like to learn how to solve or prove the above cited statement mathematically?
Thank you.
I got good reply from one of the MATH STACK USER. I studied as per his guidance about the LUCAS Theorem, I encounter the following facts with doubts and difficulties.
If we express the p (not prime) in terms of $q^x$ k where q and k are relatively primes with q is prime,. Then my example given above fails. Of course x and k are not equal to 1 simultaneously. 
With reference to the above fact, how we generalize the above fact mathematically?
Now, my second doubt/question is, why to solve my statement by Lucas Theorem? If we can do the same by Wilson’s theorem?  This is I am just guessing. I am not sure how far I am correct.  Kindly discuss, if I am wrong/correct?
If Lucas Theorem only will solve my statement,  how to encounter the fgollowing fact from Lucas theorem?
For a and q are positive integers and greater than 1, such that
$$\binom{na}{ma}$$ $\equiv 3\  $$\binom{n}{m}$  (mod p)
For every pair of integers  n greater than equal to m greater than equal to 0 with a & q are powers of the same prime p ?
I am so exited to encounter the above facts during my study on Lucas theorem to complete my statement given above. Kindly discus and thank you so much for every replier. 
 A: This is a very pretty instance of Lucas's Theorem: http://en.wikipedia.org/wiki/Lucas%27_theorem
If you write out $n$ and $p$ in base $p$ you'll see that only the "tens" digit of $n$ contributes to the product, and this digit is equivalent to $\lfloor n/p \rfloor$ modulo $p$.
Here is a more elementary argument by induction: notice that $\lfloor n/p \rfloor   - \lfloor (n-1)/p \rfloor$ is either $0$ or $1$ depending on whether $n$ is a multiple of $p$.
Therefore we want to show that the remainder of $\binom{n}{p}$ is exactly the same as $\binom{n-1}{p}$ if $n$ is not divisible by $p$, and that it is exactly $1$ higher in the case that $n$ is divisible by $p$.  Expand out:
$$\binom{n}{p}-\binom{n-1}{p} = \binom{n-1}{p-1} = \frac{(n-1)(n-2)\cdots (n-p+1)}{(p-1)(p-2)\cdots1}.$$
If $n$ is not divisible by $p$, then the numerator has one term divisible by $p$, but the denominator doesn't, so $\binom{n}{p}-\binom{n-1}{p}$ is a multiple of $p$.  This proves half of what we wanted to show.
If $n$ is divisible by $p$, then both the numerator and denominator are congruent to $(p-1)!$ mod $p$, so they cancel out to exactly $1$.  This proves the other half.
