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There are $N$ boxes, each containing at most $r$ balls. If the number of boxes containing at least $i$ balls is $N_i$ for $i=1,2,\dots,r,$ then the total number of balls contained in these $N$ boxes is exactly equal to $N_1+N_2+\dots+N_r.$

How is that exactly equal to $N_1+N_2+\dots+N_r$? I can't understand why statement containing at least and at most turn to give this exact answer!

I tested for $N=4$ and $r=3$. Suppose that box 1 has 3 balls, box 2 has 2 balls, box 3 has 1 ball and box 4 has 3 balls. So, $N_1=4,N_2=3,N_3=2$ and the conclusion holds!

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  • $\begingroup$ How many times have you counted each of the boxes containing exactly $r$ balls? $\endgroup$ – Arthur Apr 5 '14 at 8:27
  • $\begingroup$ I have shown my try. $\endgroup$ – Silent Apr 5 '14 at 8:28
  • $\begingroup$ I know, and that's appreciated. I'm trying to hint you toward why this works. So, how many of the $N_i$ terms contain in them the boxes with exactly $r$ balls? $\endgroup$ – Arthur Apr 5 '14 at 8:31
  • $\begingroup$ All the $N_i$s. $\endgroup$ – Silent Apr 5 '14 at 8:33
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Note that the number of boxes that contains exactly $j$ balls is $N_j-N_{j+1}$. Then, the total number of balls is $$\sum_{j=1}^r j(N_j-N_{j+1})=\sum_{j=1}^{r+1}N_j(j-(j-1))=\sum_{j=1}^rN_j$$ since $N_{r+1}=0$.

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  • $\begingroup$ ☺♥ Thank you so much! $\endgroup$ – Silent Apr 5 '14 at 8:42
  • $\begingroup$ How did you get $\sum_{j=1}^{r+1}N_j(j-(j-1))$? $\endgroup$ – Silent Apr 5 '14 at 9:02
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    $\begingroup$ Take for example $r=3$ : $N_1-N_2+2N_2-2N_3+3N_3-3N_4=$ $=N_1(1-0)+N_2(2-1)+N_3(3-2)+N_4(4-3)$. Note that since $N_4=0$ you can multiplicate it by whatever you want. $\endgroup$ – ajotatxe Apr 5 '14 at 9:11
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If a box contains $b$ balls, then it is counted into $N_1, N_2, \ldots, N_b$. For example a box with $3$ balls has "at least three balls", so it is counted in $N_3$, but it also has "at least two balls" and "at least one balls", so it also counted in $N_2$ and $N_1$ respectively. So you count each box as many times as is the number of balls the box contains.

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  • $\begingroup$ ☺♥ Thank you so much! $\endgroup$ – Silent Apr 5 '14 at 8:41
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It is given that $N_1$ is the number of boxes with at least $1$ ball and $N_2$ is the number of boxes with at least $2$ balls, then we can certainly conclude that the total number of boxes with exactly $1$ ball is $N_1-N_2$. Now ball contribution to the total number of balls by these $N_1-N_2$ boxes is equal to $1\times(N_1-N_2).$

Again, we have that $N_2$ is the number of boxes with at least $2$ balls and $N_3$ is the number of boxes with at least $3$ balls, then we can certainly conclude that the total number of boxes with exactly $2$ balls is $N_2-N_3$. Now ball contribution to the total number of balls by these $N_2-N_3$ boxes is equal to $2\times(N_2-N_3).$

Going on like this we can conclude that the total number of boxes having exactly $n$ balls is equal to $N_{n}-N_{n+1}$, $\forall$ $1\le n\le r-1$. Therefore ball contribution to the total number of balls by these $N_{n}-N_{n+1}$ boxes is equal to $n\times(N_{n}-N_{n+1})$.

Now the total number of boxes having at least $r$ balls is $N_r$. But, since we are provided with the constrain that each of the $N$ boxes contains at most $r$ balls, we can conclude that the total number of boxes having exactly $r$ balls is $N_r$, and ball contribution by these $N_r$ boxes is $r\times N_r$.

Therefore, total number of balls contained in these $N$ boxes is $$\sum_{i=1}^{r-1}(N_i-N_{i+1})i+rN_r=N_1+N_2+N_3+...+N_r.$$

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