Combinations With Repetition Intuition 
How many ways are there to choose a 5-letter “words” from the 26-letter English
  alphabet with repeated letters allowed, but words that are anagrams are considered the same? The words
  need not be English language words.

I believe my intuition is correct here, but upon calculating the expression, it doesn't equal to an integer. 
First we have to find all 5 letter "words" with repeating characters allowed. So we have: 
$26^5$ 
Because we have $26$ choices for each of the slots for a 5 letter word.
Next, we divide by $5!$ because we over-counted the number of words we want which excludes anagrams, and there are $5!$ ways to arrange the letters of the word.
Where am I going wrong here?
I was told that this was an application of combinations with repetition, but I don't understand why my approach is not correct.
 A: Let us count the different possibilities.

*

*If the word has $5$ different letters, then it is just $26$-choose-$5$ or $\binom{26}{5}$.


*If the word has $4$ different letters, with $1$ that is repeated, then first find the number of sets with $4$ different letters which is $\binom{26}{4}$, and now you have $4$ options to choose which letter to duplicate twice. This gives $4\binom{26}{4}$.


*If the word has $3$ different letters, you again get all the $\binom{26}{3}$ sets of $3$ different letters. Now you have two options: either you duplicate one of the three letters $3$ times - $3$ options for that, or you choose $2$ letters and duplicate each one twice - again $3$ options. In total this gives $6\binom{26}{3}$.


*If the word has $2$ different letters, you get as usual the $\binom{26}{2}$ sets of $2$ different letters and duplicate either one of the letters $4$ times - $2$ options; or duplicate one letter $2$ times, and the other $3$ times - again $2$ options. In total this is $4\binom{26}{2}$.


*Finally, if the word has just $1$ distinct letter you have $\binom{26}{1}$ options.
In total the number of different words satisfying your criterion is (see here)
$$N=\binom{26}{5}+4\binom{26}{4}+6\binom{26}{3}+4\binom{26}{2}+\binom{26}{1}=142\:506$$

Let me now suggest a way to check this via a computer. Assign a number $1-26$ to each letter in the alphabet. Convince yourself that we are looking for the number of tuples $\left(i,j,k,l,m\right)$ such that $i\leq j\leq k\leq l\leq m$. In other words, if we organize the letters in ascending order we don't overcount. The number of such tuples is
$$N=\sum_{1\leq i\leq j\leq k\leq l\leq m\leq26}1=\sum_{i=1}^{26}\sum_{j=i}^{26}\sum_{k=j}^{26}\sum_{l=k}^{26}\sum_{m=l}^{26}1=142\:506$$
according to Wolfram Alpha, confirming our previous result.
