JEE Advanced 2017 Paper 1Question 1
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, $\frac{y}{9x}=$
Permutation and Combination
Answer$5$
Solution
Concept. For $y$: choose the doubled letter, choose the $8$ other letters, arrange $10$ letters with one pair. Formulas used. (1) $x=10!$(2) Arrangements with one pair: $\frac{10!}{2!}$Step 1 - $x$. All $10$ letters once: $10!$. Step 2 - Doubled letter. $10$ ways. Step 3 - Other 8 letters. From the remaining $9$: ${}^9C_8=9$. Step 4 - $y$. $10\cdot9\cdot\frac{10!}{2}=45\cdot10!$. Step 5 - Ratio. $\frac{y}{9x}=\frac{45\cdot10!}{9\cdot10!}=5$. Answer: 5