Total number of possibilities I have 3 buckets, 1st bucket has 5 red balls, 2nd bucket has 3 green balls and 3rd one has 2 blue balls. so I have total 10 balls in 3 buckets. 

I need to know, what are the possible combination available if I need to pick up 4 balls( irrespective of colors) from 3 buckets. i.e I can pick up 4 from bucket 1 , or I can pick them up from different buckets as well.i.e 2 red, 1 green, 1 blue or 4 red or 2 green and 2 blue or 2 blue and 2 red...etc..what will be the mathematical formulation
We assume that balls of the same colour are identical. Let $r$ be the number of red we pick, $g$ the number of green, and $b$ the number of blue. We want to find the number of solutions of the equation $r+g+b=4$ in non-negative integers.

This would be taken care of by a general formula ("Stars and Bars") if there were $4$ or more balls in each bucket. However, there are only $2$ blue balls, so for example the solution $(0,1,3)$ is not possible. 

I do not know whether you are expected to use a "general" procedure, or just plain count. We will do the latter, in a reasonably organized manner. 

Divide into cases. 

4 red: There is only $1$ way to do this.

3 red: We can pick a green, or a blue, $2$ ways.

2 red: We pick $2$ green, or $2$ blue, or $1$ of each, a total of $3$ ways.

1 red: We pick $3$ green, or $2$, or $1$, and the rest blue, a total of $3$ ways.

0 red: We pick $3$ green, or $2$, and the rest blue, a total of $2$ ways.

Add up. 

Remark: Since the blues are fewest, it is obviously more efficient to let the cases be $2$ blue, $1$ blue, $0$ blue. You should do it that way, in order to make the solution "your own."
Since you don't care about colors, you will need to find possible combinations of choosing 4 balls from 10 balls.
$10!/6!4!=210$ combinations