[Combinatorics]
Problem: The ARMLLexicon consists of 10 letters: $\{A, R, M, L, e, x, i, c, o, n\}$. A palindrome is an ordered list of letters that read the same backwards and forwards; for example, MALAM, n, oncecno, and MoM are palindromes. Compute the number of 15-letter palindromes that can be spelled using letters in the ARMLLexicon, among which there are four consecutive letters that spell out $A R M L$.

Solution: Any 15-letter palindrome is determined completely by its first 8 letters, because the last 7 letters must be the first 7 in reverse. Such a palindrome contains the string $A R M L$ if and only if its first 8 letters contain either $A R M L$ or $L M R A$. (The string $A R M L$ cannot cross the middle of the palindrome, because the 7th and 9th letters must be the same.) It therefore suffices to count the number of 8-letter strings consiting of letters in the ARMLLexicon that contain either ARML or LMRA.

There are 5 possible positions for $A R M L$, and likewise with $L M R A$. For each choice of position, there are four remaining letters, which can be any letter in the ARMLLexicon (here, $W, X, Y$, and $Z$ are used to denote arbitrary letters that need not be distinct). This leads to the following table:



| Word | Num. Possibilities |
| :---: | :---: |
| ARMLWXYZ | $10^{4}$ |
| WARMLXYZ | $10^{4}$ |
| WXARMLYZ | $10^{4}$ |
| WXYARMLZ | $10^{4}$ |
| WXYZARML | $10^{4}$ |
| LMRAWXYZ | $10^{4}$ |
| WLMRAXYZ | $10^{4}$ |
| WXLMRAYZ | $10^{4}$ |
| WXYLMRAZ | $10^{4}$ |
| WXYZLMRA | $10^{4}$ |

This gives $10 \cdot 10^{4}$ possible words, but each word with two of ARML or LMRA (e.g., ARMLARML or $A A R M L M R A$ ) is counted twice. There are four words with two of $A R M L$ or $L M R A$ that use all 8 letters, and four possible types of words that use 7 of the 8 positions and leave one "free space". This leads to the following table:

| Word | Num. Possibilities |
| :---: | :---: |
| ARMLARML | 1 |
| LMRALMRA | 1 |
| ARMLLMRA | 1 |
| LMRAARML | 1 |
| ARMLMRAW | 10 |
| LMRARMLW | 10 |
| WARMLMRA | 10 |
| WLMRARML | 10 |

Thus the total number of desired words is $10 \cdot 10^{4}-4 \cdot 10-4 \cdot 1=\mathbf{9 9 9 5 6}$.

Final answer: 99956