Distribute $a + b + c$ distinct balls into boxes $A, B, C$ such that $a$ balls, $b$ balls and $c$ balls go to boxes $A, B, C$ 
Distribute $a + b + c$ distinct balls into boxes $A, B, C$ such that $a$ balls, $b$ balls and $c$ balls go to boxes $A, B, C$, respectively, and show that the number of ways to do this is $\frac{(a+b+c)!}{a!b!c!}$

I want to do the special case just to see if I am thinking correctly about this. Please, see if that makes sense.
Show the number of ways to distribute $10$ distinct balls into boxes $A, B, C$ is $\frac{(5 + 3 + 2)!}{5!2!3!}$ if exactly five balls go to box $A$ exactly three balls go to box $B$ and exactly two balls go to box $C$. 
There are $10$ ways to choose the first ball going into the box $A$, there are $9$ ways to choose the second ball going into the box $A$...in all there are $10 \cdot 9 \cdot 8 \cdot 7 \cdot 6$ ways to put five balls into the box $A.$ For each such placement of five balls in $A$, we can place three balls in $5 \cdot 4 \cdot 3$ ways. Finally, there are two ways to place the last two balls in $C.$ In total, there are $10!$ ways to place ten distinct balls in the boxes $A, B, C.$ But we have overcounted by the factor of $5!3!2!$ so we divide them out: $\frac{(5 + 3 + 2)!}{5!2!3!}.$

You say that "in all there are $10⋅9⋅8⋅7⋅6$ ways to put five balls into the box A". That is only true if the order in which we put the balls into the box matters. Clearly, it doesn't over here thus the correct computation would be $\displaystyle\binom{10}{5},$ that is $\dfrac{10⋅9⋅8⋅7⋅6}{5!}$ since we divide by the number of ways the $5$ balls could be ordered. Having selected $5$ balls for box A, you can select $3$ balls for box B in $\displaystyle\binom{5}{3}$ ways and $2$ balls for box C in $\displaystyle\binom{2}{2}=1$ way. Since these choices are independent, the total number of ways to assign $10$ balls to $3$ boxes subject to those conditions is $\displaystyle\binom{10}{5}\cdot\binom{5}{3}\cdot\binom{2}{2}= \displaystyle\binom{10}{5,3,2}=\dfrac{10!}{5!\cdot3!\cdot2!}$, if you're familiar with multinomial coefficients.
