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 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
  1. 45
  2. 415
  3. 315
  4. 215

Solution

$\begin{aligned} & \text { Letters are A, B, C, D, E, F, G, H, I, J } \\ & \text { Number of words that can be formed by } \\ & 10 \text { letters }=10!\times{ }^{10} \mathrm{C}_{10} \\ & \therefore \quad x=10! \end{aligned}$
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 $\begin{aligned} & ={ }^{10} \mathrm{C}_2 \times{ }^8 \mathrm{C}_6 \times \frac{10!}{2!\times 2!} \\ \therefore \quad y & ={ }^{10} \mathrm{C}_2 \times{ }^8 \mathrm{C}_6 \times \frac{10!}{2!\times 2!} \\ \therefore \quad \frac{y}{x} & =\frac{{ }^{10} \mathrm{C}_2 \times{ }^8 \mathrm{C}_2 \times \frac{10!}{2!\times 2!}}{10!} \\ & =\frac{{ }^{10} \mathrm{C}_2 \times{ }^8 \mathrm{C}_6}{2!\times 2!} \\ & =\frac{45 \times 28}{4} \\ & =315\end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 1)

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