Ratio of Word Permutations with Repeated Letters

To find the value of $\frac{y}{9x}$, we evaluate $x$ and $y$ step-by-step:

1. Calculate $x$ (Number of 10-letter words with no repeated letters): There are 10 distinct letters available: $\{\text{A, B, C, D, E, F, G, H, I, J}\}$. Since the words are of length 10 and no letter is repeated, each letter must be used exactly once. The total number of such arrangements is given by: $x = 10!$

2. Calculate $y$ (Number of 10-letter words with exactly one letter repeated twice and no other repetitions): To form a 10-letter word with this property:

   • Step 1: Choose $1$ letter out of $10$ available letters to be repeated twice. The number of ways to do this is: $\binom{10}{1} = 10$
   • Step 2: Choose $8$ distinct letters from the remaining $9$ letters to fill the remaining $8$ positions in the word. The number of ways to do this is: $\binom{9}{8} = 9$
   • Step 3: Arrange the selected letters (which consist of 1 letter appearing twice and 8 distinct letters appearing once, making a total of 10 letters) to form a word. The number of permutations is: $\frac{10!}{2!}$

   Multiplying these together gives the total count $y$: $y = \binom{10}{1} \times \binom{9}{8} \times \frac{10!}{2!} = 10 \times 9 \times \frac{10!}{2} = 45 \times 10!$

3. Compute the ratio $\frac{y}{9x}$: $\frac{y}{9x} = \frac{45 \times 10!}{9 \times 10!} = \frac{45}{9} = 5$

Thus, the value of $\frac{y}{9x}$ is $5$.