# 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, $\dfrac{y}{9x}$ is equal to

Last updated date: 28th Mar 2023

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Hint: First we will find the number of combinations of words formed when no letter is repeated and then we will find the number of combinations of words where exactly one letter is repeated twice and finally we will find $\dfrac{y}{9x}$.

Complete step-by-step answer:

Given that;

The words of length 10 are formed using letter A, B, C, D, E, F, G, H, I, J.

Let us assume x be the number of words where no letter is repeated.

Choosing the word where no letter is repeated can be done in 10! ways.

$\therefore $ we get x = 10!

Let us assume y be the number of words where exactly one letter is repeated twice.

Since, one letter is repeated.

So, we can choose one letter out of 10 letters and can be done in ${}^{10}{{C}_{1}}$ ways, then place them on any two places, which can be done in ${}^{10}{{C}_{2}}$ ways.

So,

$\begin{align}

& y={}^{10}{{C}_{1}}\times {}^{10}{{C}_{2}}\times {}^{9}{{C}_{8}}\times 8! \\

& y=\dfrac{10!}{1!\left( 10-1 \right)!}\times \dfrac{10!}{2!\left( 10-2 \right)!}\times \dfrac{9!}{8!\left( 9-8 \right)!}\times 8! \\

& y=\dfrac{10\times 9!}{9!}\times \dfrac{10\times 9}{2}\times \dfrac{9\times 8!}{8!}\times 8! \\

& y=10\times 45\times 9\times 8! \\

\end{align}$

Now, we have all the required data to solve this problem.

We have x = 10!

And y = $10\times 45\times 9\times 8!$

Then, $\dfrac{y}{9x}=?$ ………………. (1)

On substituting the value of x and y in equation (1), we get;

\[\begin{align}

& \dfrac{y}{9x}=\dfrac{10\times 45\times 9\times 8!}{9\times 10!} \\

& =\dfrac{{10}\times 45\times{9}\times{8!}}{{9}\times{10}\times 9\times {8!}} \\

& =\dfrac{45}{9} \\

& =5 \\

\end{align}\]

So, the answer is 5.

Note: Please note that if repetition is not allowed then the total possibility will be n! but the repetition is allowed the total possibility will be ${{\left( n \right)}^{n}}$. Here in this question repetition is not allowed thus, we have the value of x = 10!.

Complete step-by-step answer:

Given that;

The words of length 10 are formed using letter A, B, C, D, E, F, G, H, I, J.

Let us assume x be the number of words where no letter is repeated.

Choosing the word where no letter is repeated can be done in 10! ways.

$\therefore $ we get x = 10!

Let us assume y be the number of words where exactly one letter is repeated twice.

Since, one letter is repeated.

So, we can choose one letter out of 10 letters and can be done in ${}^{10}{{C}_{1}}$ ways, then place them on any two places, which can be done in ${}^{10}{{C}_{2}}$ ways.

So,

$\begin{align}

& y={}^{10}{{C}_{1}}\times {}^{10}{{C}_{2}}\times {}^{9}{{C}_{8}}\times 8! \\

& y=\dfrac{10!}{1!\left( 10-1 \right)!}\times \dfrac{10!}{2!\left( 10-2 \right)!}\times \dfrac{9!}{8!\left( 9-8 \right)!}\times 8! \\

& y=\dfrac{10\times 9!}{9!}\times \dfrac{10\times 9}{2}\times \dfrac{9\times 8!}{8!}\times 8! \\

& y=10\times 45\times 9\times 8! \\

\end{align}$

Now, we have all the required data to solve this problem.

We have x = 10!

And y = $10\times 45\times 9\times 8!$

Then, $\dfrac{y}{9x}=?$ ………………. (1)

On substituting the value of x and y in equation (1), we get;

\[\begin{align}

& \dfrac{y}{9x}=\dfrac{10\times 45\times 9\times 8!}{9\times 10!} \\

& =\dfrac{{10}\times 45\times{9}\times{8!}}{{9}\times{10}\times 9\times {8!}} \\

& =\dfrac{45}{9} \\

& =5 \\

\end{align}\]

So, the answer is 5.

Note: Please note that if repetition is not allowed then the total possibility will be n! but the repetition is allowed the total possibility will be ${{\left( n \right)}^{n}}$. Here in this question repetition is not allowed thus, we have the value of x = 10!.

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