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**Grandad** Hello everyoneMay I amplify Plato's solution a little, and query one of the results.

Case I. This is where we select 1 of each of the 4 different letters; so that's $\displaystyle {^4C_4}=1$ selection and $\displaystyle 4!=24$ arrangements of these 4 different letters.

Case II. One letter repeated; for example: e, e, n, s.

There are 3 ways of selecting which letter is to be repeated, and then $\displaystyle {^3C_2} = 3$ ways of selecting which two of the remaining 3 different letters will accompany the repeated letter. So that's $\displaystyle 3\times 3 =9$ selections.

We can then arrange 4 letters which contain one pair of repeated letters in $\displaystyle \frac{4!}{2!}=12$ ways.

Case III. Two letters repeated: e, e, n, n, for instance. I think that's just $\displaystyle {^3C_2} = 3$ ways of selecting which 2 of the 3 repeated letters we'll use. So I get a different answer here.

Then I agree with the answer of $\displaystyle \frac{4!}{(2!)^2}=6$ arrangements of 4 letters with 2 repeated pairs.

Case IV. I think this means 1 letter repeated 3 times. So that's the letter e with one of the remaining 3. That's 3 choices and $\displaystyle \frac{4!}{3!}=4$ arrangements of 4 items with a group of 3 'like' items.

Case V. This means 1 letter repeated 4 times. So that's 1 selection (all e's) and 1 arrangement.

So, when we add up, we get 17 selections (combinations) and 47 arrangements (permutations).

Grandad