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, $\frac{y}{9x}=$

Answer 1

Letters are $A,B,C,D,E,F,G,H,I,J$
Number of words that can be formed by $10$ letters is $=10!\times {}^{10}{C}_{10}=10!$
$\therefore x=10!$

Now, for repetition of one letter :
The particular letter can be selected in ${}^{10}{C}_1$ ways which is used twice in the word.
And rest of the $8$ letters can be selected from $9$ letters in ${}^9{C}_8$ ways.
Hence number of word than can be formed $={}^{10}{C}_1\times {}^9{C}_8\times \frac{10!}{2!}$
$\therefore y={}^{10}{C}_1\times {}^9{C}_8\times \frac{10!}{2!}\Rightarrow \frac{y}{9x}=\frac{{}^{10}{C}_1\times {}^9{C}_8\times \frac{10!}{2!}}{9\times 10!}=5$