Is this a combinations or permutations problem? If you are trying to partition a message of length 50 into 7 partitions, where there is atleast one letter in each, would that be a combinations or permutations problem?
 A: To create 3 partitions (according to initial problem statement):

*

*Let the partition indices for the message $S[1 \ldots 50]$ be $i$ and $j$, with $2 \leq i<j \leq 50$, so that the three partitions are $S[1\ldots i-1], S[i\ldots j-1], S[j\ldots 50]$, s.t., total number of ways to create such partitions
$=\sum\limits_{i=2}^{49}\sum\limits_{j=i+1}^{50}1$
$=\sum\limits_{i=2}^{49}(50-i)=48.50-49.50/2+1=1176$.


*Thinking in another way, we need to select 2 indices $i$ and $j$, with $i<j$, out of 50 indices from $1 \ldots 50$, s.t., left of $i$ is the first partition, from $i$ to $j-1$ is the second partition and from $j$ to 50 is the 3rd partition.


*Now, each partition must have at least one letter, so $i \geq 2$, i.e., the problem reduces to selecting 2 indices from $2 \ldots 50$, i.e., out of 49 indices that can be done in ${49}\choose {2}$ $=48.49/2=1176$ ways. This is a combination problem, since the letters in the message can't be rearranged inside a partition.
To create 7 partitions (the problem statement got changed to this later):
the same logic can be used and in this case it can be done in ${49}\choose {6}$ ways.
Generalizing, for length $n$ message and to create $k$ partitions so that each partition contains at least one alphabet, it can be done in ${n-1}\choose{k-1}$ ways, since we need to just select $k-1$ split points from $n-1$ possible candidate indices.
