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Question

Words of length $10$ are formed using the letters A, B, C, D, E, F, G, H, I, J. Let $x$ be the number of such words where no letter is repeated; and let $y$ be the number of such words where exactly one letter is repeated twice and no other letter is repeated. Then

$$\frac{y}{9 \times x}=?$$

My Approach

$x=10!$ as letters are not allowed to be repeated and we have 10 letters available , so word of length $10=10 \times 9 \times 8 ..1=10!$


Now calculating $y$,

We have to select which letters are being repeated ,so

$$\binom{10}{1}=10 $$ now letters which will be repeated are already selected. So, possible # of words=$$\binom{10}{1} \times \frac{10!}{2!}$$ as $2$ letters among $10$ are repeated.

so $$\frac{y}{9 \times x}=\frac{5}{9}$$

But the answer is $5$. What am I doing wrong? Please help!

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As regards $y$, note that we have to choose also the letter to be excluded. Hence it should be $$y=\underbrace{10}_{\text{repeated letter}}\times \underbrace{9}_{\text{excluded letter}} \times \underbrace{\frac{10!}{2!}}_{\text{anagrams}}$$

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