Combinatoric putting balls in boxes 9 tennis balls, which are numbered 1 through 9 inclusively, will be put into the cylindrical transparent boxes. 2 balls into box A, 3 balls into box B, 4 balls into box C. Further, the balls must be in descending order in the boxes.
In how many different ways can the 9 balls put into the boxes?
I simply think C(9,2)*C(7,3)*C(4,4), but seems absolutely wrong. I really want to comprehend the kind of questions. Could you explain the answer as explaining to a stupid?

 A: Your answer is correct. Here is the explanation of why.
Please realize this fact:
Once we put any amount of balls into each box, there is only ONE way to satisfy the order rule of decending.
For example, if ball 1,3,5 are in box B, there is only one decending order  5,3,1.
Therefore, we just need to take care how these 9 balls are assigned to each  box :
$(C^9_2 \cdot 1 )(C^7_3 \cdot 1)(C^4_4\cdot 1)$ .
A: Your answer is absolutely correct . Whatever balls you choose for the boxes , they can always be arranged the order you wish , so the number of ways is equal to "selecting" the balls required .
Therefore , your answer ,
$C^9_2 * C^7_3 * C^4_4$ is correct.
A: For the first spot you want to choose $2$ balls out of $9$ to go into the first container. Since order doesn't matter we can choose any two balls, and order doesn't matter we will use combinations. Thus, $\binom{9}{2}$.
Now we have $7$ balls left. So we choose $3$ balls to go into the second container, $\binom{7}{3}$. Do the same for the last container, $\binom{4}{4}$.
Your answer is correct! $\binom{9}{2}\cdot \binom{7}{3} \cdot \binom{4}{4}$
Context: $\binom{9}{2}$ is equivalent to $C(9,2)$
