Can someone please explain this. How many 4 letter words are there, when order doesn't matter and letters can be repeated ? IF I do in one approach I get $\frac{26^4}{4!} $ (26 letters for each position and then divide by 4! as the position didn't matter).
Is this logic correct ? I Am not sure because, the course video I am watching gives a different logic as below.
by composition : $a+b+c+...+z = 4$ ; therefore can use the $n+k-1\choose k-1$ formula(stars and bars), where $n$ is $4$ and $k$ is $26$.This approach also seems correct but the final answers for approach 1 and 2 are different. Thanks in advance
You are thinking in the right direction, but your answer is a little off. Not every combination is repeated $4!=24$ times, so you can't just divide by $24$ and call it a day. For example, the combination AAAA is only counted once from the $26^4$ part of your answer, but you are dividing by $24$, so you will actually get $\frac{1}{24}$ of a way to get the combination AAAA, which is clearly incorrect. What you might want to do is to split the problem into four cases(the stars and bars approach is probably an easier way but I'm trying to do an approach similar to yours):