The number of different permutations of lette | vedclass

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(C) The word '$REPETITION$' consists of $10$ letters: $R, E, P, E, T, I, T, I, O, N$. The distinct letters are $R, E, P, T, I, O, N$ ($7$ distinct let

Vedclass · Accepted Answer

$1398$

Vedclass · Answer

(C) The word '$REPETITION$' consists of $10$ letters: $R, E, P, E, T, I, T, I, O, N$. The distinct letters are $R, E, P, T, I, O, N$ ($7$ distinct letters). The repeated letters are $E, T, I$ (each appearing twice). We need to form permutations of $4$ letters. The cases are: $(i)$ All $4$ letters are distinct: We choose $4$ letters from $7$ distinct letters and arrange them in $4!$ ways. Number of ways $= {}^{7}C_{4} 	imes 4! = 35 	imes 24 = 840$. (ii) $2$ letters are the same (one pair) and $2$ are distinct: We choose $1$ pair from $3$ available pairs $(E, T, I)$ and $2$ letters from the remaining $6$ distinct letters. Then we arrange them. Number of ways $= {}^{3}C_{1} 	imes {}^{6}C_{2} 	imes \frac{4!}{2!} = 3 	imes 15 	imes 12 = 540$. (iii) $2$ pairs of letters: We choose $2$ pairs from $3$ available pairs. Then we arrange them. Number of ways $= {}^{3}C_{2} 	imes \frac{4!}{2! 	imes 2!} = 3 	imes 6 = 18$. Total permutations $= 840 + 540 + 18 = 1398$. Hence,option $(C)$ is correct.

The number of different permutations of letters that can be formed by taking $4$ letters at a time from the letters of the word '$REPETITION$' is

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