Suppose the following list: $$ \begin{matrix} j_1 & j_2 & j_3 & j_4\\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 2 \\ 1 & 1 & 1 & 3 \\ 1 & 1 & 2 & 1 \\ 1 & 1 & 2 & 2 \\ & \vdots & \\ 3 & 3 & 3 & 2 \\ 3 & 3 & 3 & 3 \end{matrix} $$ In this list, lets group repeated entries in the columns such that: $$ \begin{matrix} j_1 & j_2 & j_3 & j_4\\ \{1 & 1 & 1 & 1 \}\\ \{1 & 1 & 1\} & \{2\} \\ \{1 & 1 & 1\} & \{3\} \\ & \vdots & \\ \{2 & 2 & 2 & 2\} \\ & \vdots & \\ \{3\} & \{1 & 1 & 1\} \\ \{3 & 3\} & \{2 & 2\} \\ & \vdots & \\ \{3 & 3 & 3 & 3\} \end{matrix} $$
Is there a way to calculate how many entries the subgroups are equal to a certain number? Notice that the subgroup $\{ j_1 j_2 j_3 j_4 \}$ counts 3. On the other hand, it is not that obvious to calculate for the other groups.
Is there a generalized way to calculate this if I have $j_1, j_2, \dots, j_n $ ?
Answering directly for the generalised case:
Let's denote the arrangement as:
$$\_ \ \_ \ \_ \ .....(n \ times)$$
So let's say that each $\_$ is can be filled with $k$ options. ($k=3$ in the example in the question)
Now if we need to calculate for $r$ repetitions of a option from the $k$ options, ($r=2$ in the example in the question)
First choose an option from $k$ : $\binom{k}{1}$ ways
and then choose $r$ places to place this option: $\binom{n}{r}$
Now look at the other $n-r$ places left, where you can place different characters from $k-1$ options : $\binom{k-1}{n-r} (n-r)!$ ways.
So the total ways:
$$\binom{k}{1}\binom{n}{r}\binom{k-1}{n-r}(n-r)!$$
(Place respective bounds on r,k)
So for your example in the question:
$n=3$, we have total of 3 options: 1, 2, 3 ($k=3$)
So for 3 repetitions, ($r=3$)
$$=\binom{3}{1}\binom{3}{3}\binom{2}{0}(0!)=3$$
For 2 repetitions ($r=2$)
$$=\binom{3}{1}\binom{3}{2}\binom{2}{1}(1)!=18$$
Hope that helped...... :-)