# Words of length $10$ are formed using the letters A, B, C, D, E, F, G, H, I, J

## 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!

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