Repel Kepler Peel

Let's solve this problem as if the name in question were EUCLID first, then figure out how to deal with the duplicate letter.

I know of no formula for this, so let's start small and see what pattern emerges:

If the source word had 1 letter ("A"), we could clearly make 1 word ("A").

If it had 2 letters ("AB"), we could make 4 ("A", "B", "AB", "BA").

For 3, we have: 3P1=3 of length 1, 3P2=6 of length 2 and 3P3=6 of length 3. 3 + 6 + 6 = 15.

For 4, 4P1 + 4P2 + 4P3 + 4P4 = 4 + 12 + 24 + 24 = 64.

So far, 1, 4, 15, 64. If we continue in this fashion we will see a pattern emerge: when adding the nth letter, we start from the previous number, add one, then multiply by n. For 5 we will have 325, and for 6 (as in EUCLID), 1956 (which I verified longhand, to be safe).

Now we must deal with that pesky duplicate E in KEPLER. How many have we overcounted? It turns out that every word containing any E in it has been counted exactly twice:

• for the words with only one E in them, they were counted once for each of the two Es in the source word.
• whereas for the words with two Es, they were counted once for each of the two arrangements those two Es can be in.

So everything was double-counted except the words containing no Es at all. Since those are all the possible words using the letters KPLR, there must be 64 of them as we worked out above (how convenient!). So 1956 - 64 = 1892 of them are our doubles, representing 946 unique overcounted words.

We can take either 946 + 64 or 1956 - 946 to get our answer, 1010.

(Verified with a simple computer program to generate and unique-ify the actual words.)

To solve this problem we have to find the number of distinct words for different lengths.

$\textbf{Formulas: }$ $n$ is the number of choices, $r$ is how many we choose. $\begin{array}{cc} & _{\text{Permutation: Order matters}} &\quad _{\text{Combination: Order doesn't matter}} \\ & P(n,r) = \frac{n!}{(n-r)!} & \quad C(n,r) = \frac{n!}{r!(n-r)!} \end{array}$

$\textbf{Length 6: }$ First we find the number of combinations that we can have of the two E's in the word. Then we multiply the number of combinations by the number of permutations we can have for the remaining four letters. Another way to find this case is to take the number of permutations as if all 6 letters are unique, then divide by 2 since each word will be a duplicate. However, that way will not work when the length is less than 6.

$C(6,2) = 15 \quad P(4,4) = 24 \quad \Rightarrow \quad 15\cdot24 = \mathbf{360} \\$

$\textbf{Length 5: }$ For this case and the rest, we further split them each into 2 mutually exclusive cases. One where the word contains 2 E's and the other when the word contains 1 E. For the first situation, we calculate it similarly as the case with length 6. For the second situation, we can go ahead and do a permutation since we don't have to worry about duplicates. Then we just add the number from each case.

$C(5,2) = 10 \quad P(4,3) = 24 \quad \Rightarrow \quad 10\cdot24 = 240 \\ P(5,5) = 120 \quad \quad \Rightarrow \quad 240 +120 = \mathbf{360} \\$

$\textbf{Length 4: }$

$C(4,2) = 6 \quad P(4,2) = 12 \quad \Rightarrow \quad 6\cdot12 = 72 \\ P(5,4) = 120 \quad \quad \Rightarrow \quad 72 +120 = \mathbf{192} \\$

$\textbf{Length 3: }$

$C(3,2) = 3 \quad P(4,1) = 4 \quad \Rightarrow \quad 3\cdot4 = 12 \\ P(5,3) = 60 \quad \quad \Rightarrow \quad 12 +60 = \mathbf{72} \\$

$\textbf{Length 2: }$ For this case, the first situation is a bit different since there is only 1 combination for 2 E's and we don't have any remaining letters to choose. But the calculation would still turn out the same if we chose to continue with the method from the previous cases.

$C(2,2) = 1 \quad P(4,0) = 1 \quad \Rightarrow \quad 1\cdot1 = 1 \\ P(5,2) = 20 \quad \quad \Rightarrow \quad 1 + 20 = \mathbf{21} \\$

$\textbf{Length 1: }$ For this last case, it is obvious that we can 5 different words. But the formula for permutations would still work if we chose to use it.

$P(5,1) = \mathbf{5} \\$

Now we simply add up the total from each case and we will have the number of unique words with length 1- 6.

$360 + 360 + 192 + 72 + 21 + 5 = \fbox{1010 } \\$