Words of length 10 are formed by using the letters A, B, C, D, E, F, G, H, I, J. Let \(x\). be number of such words where no letter is repeated and \(y\) be number of such words where exactly two letters are repeated twice and no other letter is repeated, then the value of \(\frac{y}{x}\) is

Letters are A, B, C, D, E, F, G, H, I, J
Number of words that can be formed by
\( 10 \text { letters }=10!\times{ }^{10} \mathrm{C}_{10} \)
\( \therefore x=10!\)
Now, for repetition of two letters.
Two letters can be selected in \({ }^{10} \mathrm{C}_2\) ways which are used twice in the word and remaining 6 letters can be selected from 8 letters in \({ }^8 \mathrm{C}_6\) ways.
Hence, Number of words can be formed
\(={ }^{10} \mathrm{C}_2 \times{ }^8 \mathrm{C}_6 \times \frac{10!}{2!\times 2!} \)
\( \therefore y ={ }^{10} \mathrm{C}_2 \times{ }^8 \mathrm{C}_6 \times \frac{10!}{2!\times 2!} \)
\( \therefore \frac{y}{x} =\frac{{ }^{10} \mathrm{C}_2 \times{ }^8 \mathrm{C}_2 \times \frac{10!}{2!\times 2!}}{10!} \)
\( =\frac{{ }^{10} \mathrm{C}_2 \times{ }^8 \mathrm{C}_6}{2!\times 2!} \)
\( =\frac{45 \times 28}{4} \)
\( =315\)