We have $n_1$ number of balls each labelled by the number $1$, $n_2$ number of balls each labelled $2$,..., $n_k$ number of balls each labelled $k$. In how many ways can we permute them in $r$ number of boxes assuming each box can hold exactly $1$ ball.


Finding permutations for $n_1$ + $n_2$ + ... + $n_k$ = $r$ would be the same as finding permutations with repetition. Moreover if we have $n_1$ + $n_2$ + ... + $n_k$ < $r$ we may introduce $r$ - ($n_1$ + $n_2$ + ... + $n_k$) balls of another type and find permutations with repetition for $r$ total balls. So finding permutations for $n_1$ + $n_2$ + ... + $n_k$ > $r$ would be enough to answer this question.

  • $\begingroup$ Hint: How many ways are there to permute $N$ objects in $r$ ways given that there are $n_k$ repetitions in the $N$ objects for each $k$ ? $\endgroup$ – Prasun Biswas Oct 8 '17 at 10:31
  • $\begingroup$ @HDatta - I didn't get the significance of adding $r$ - ($n_1$ + $n_2$ + ... + $n_k$) balls when total balls $< r$. Would you pls write down your solution for this case! $\endgroup$ – KGhatak Oct 14 '17 at 18:28
  • $\begingroup$ @KGhatak By adding extra balls the total no. of balls become $r$. These new balls just represent empty boxes. So permutation with repetition for $r$ balls should give us the solution for this case. $\endgroup$ – HDatta Oct 14 '17 at 18:32
  • $\begingroup$ @HDatta - So, are you saying # of ways arranging 2 red and 3 blue balls in 5 boxes is same as the # ways we can arrange a red and a blue balls in 5 boxes? Not sure if I'm missing anything! maybe writing exact answer might eliminate the confusion!! $\endgroup$ – KGhatak Oct 14 '17 at 18:54
  • $\begingroup$ @KGhatak - The balls added are of another kind acting just as empty sapces. For example if we want to find all arrangements for $n_1$ = $2$, $n_2$ = $3$ and $r$ = $7$ then we may introduce $n_3$ = $2$ balls of another kind and find permutations for $r$ total balls with repetitions. $\endgroup$ – HDatta Oct 15 '17 at 3:15

We have $ \sum_{i=1}^{k} n_i \leq r$. There are $ \binom{r}{n_1}$ ways to place the $1$'s. Then there are $ \binom{r-n_1}{n_2}$ ways to place the $2$'s. ... multiplying these binomial coefficients together gives the multinomial coeffiecient

\begin{eqnarray*} \binom{r}{n_1 , n_2, \cdots , n_k,r-\sum_{i=1}^{k} n_i}. \end{eqnarray*}

  • $\begingroup$ I didn't specify that n_1 + n_2 +...+n_k add up to r. They could be equal greater than or lesser than r. $\endgroup$ – HDatta Oct 7 '17 at 14:40
  • $\begingroup$ Thank you for responding earlier. I know that I am late but I still updated my question to clarify things a bit. Could you please have a look at this question again? $\endgroup$ – HDatta Oct 14 '17 at 16:55

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