How to distribute 5 cards among 3 people $A,B,C$ so that none of the person have empty hands 
How to distribute 5 cards among 3 people $A,B,C$ so that none of the person have empty hands?

My try:
5 cards can be distributed among $A,B,C$ in $2,2,1$ ways.
Now there are three ways of arranging $2,2,1$ .
Once I decide whether A or B or C is 2 or 1 we have $\binom 5 2$ ways of distributing 2 cards to the person who has 2 cards ,$\binom 3 2$ ways of distributing 2 cards to another  person who has 2 cards 
Thus the answer is $\binom 5 2\times \binom 3 2\times 3$
Is the answer correct??
Please correct if not .

That's correct, but you are forgetting the case $1,1,3$. We can permute $1,1,3$ in $3$ ways, select 3 cards in $\binom53$ ways, the next 2 cards in $\binom21,\binom11$ ways respectively. The total ways are $3\binom53\binom21\binom11=60$. Add to it $3\binom52\binom32=90$ to get a total of $150$.
