The user wants me to solve a permutation and combination problem.

Question Breakdown:

1. We are forming words of length 10.
2. The alphabet consists of 10 distinct letters: A, B, C, D, E, F, G, H, I, J.
3. We need to find x, the number of 10-letter words with no repeated letters.
4. We need to find y, the number of 10-letter words where exactly one letter is repeated twice, and no other letter is repeated.
5. Finally, we need to calculate the value of y / (9x).

Step 1: Calculate the value of x

• x represents the number of 10-letter words formed using the 10 distinct letters A, B, C, D, E, F, G, H, I, J, where no letter is repeated.
• This is equivalent to finding the number of permutations of 10 distinct items, which is 10!.
• So, x = 10!.

Step 2: Calculate the value of y

• y represents the number of 10-letter words where exactly one letter is repeated twice and no other letter is repeated. This means the word is formed using 9 distinct letters, with one of them appearing twice.

• We can construct such a word in the following steps:

  1. Choose the 9 distinct letters to be used from the available 10 letters. The number of ways to do this is ¹⁰C₉. $¹⁰C₉ = {10! \over {9!(10-9)!}} = 10$
  2. From these 9 chosen letters, select one letter to be repeated twice. The number of ways to do this is ⁹C₁. $⁹C₁ = 9$
  3. Arrange the 10 letters. We now have 10 letters to arrange in a 10-letter word. This collection of letters contains one letter that appears twice and 8 other letters that appear once. The number of distinct permutations of these 10 letters is given by 10! / 2!. $\text{Arrangements} = {10! \over 2!}$

• The total number of such words, y, is the product of the number of ways in each step: $y = (¹⁰C₉) \times (⁹C₁) \times \left({10! \over 2!}\right)$ $y = 10 \times 9 \times {10! \over 2}$ $y = 90 \times {10! \over 2}$ $y = 45 \times 10!$

• Alternative method for calculating y:

  1. Choose 2 positions for the repeated letter from the 10 available positions: ¹⁰C₂ ways.
  2. Choose which letter (out of 10) will be repeated: ¹⁰C₁ ways.
  3. Fill the remaining 8 positions with 8 distinct letters chosen from the remaining 9 letters. This is a permutation ⁹P₈. $y = ¹⁰C₂ \times ¹⁰C₁ \times ⁹P₈$ $y = \left({10 \times 9 \over 2}\right) \times 10 \times {9! \over (9-8)!}$ $y = 45 \times 10 \times 9!$ $y = 45 \times (10 \times 9!) = 45 \times 10!$ Both methods yield the same result for y.

Step 3: Calculate the value of y / (9x)

• Now we substitute the expressions we found for x and y into the required ratio. $x = 10!$ $y = 45 \times 10!$
• The ratio is: ${y \over {9x}} = {45 \times 10! \over 9 \times 10!}$
• The 10! terms in the numerator and denominator cancel out. ${y \over {9x}} = {45 \over 9}$ ${y \over {9x}} = 5$
• The final answer is 5.