Words of length 10 are formed by using the le | vedclass

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(C) The set of letters is $\{A, B, C, D, E, F, G, H, I, J\}$,which contains $10$ distinct letters. For $x$,we form words of length $10$ using all $10$

Vedclass · Accepted Answer

$315$

Vedclass · Answer

(C) The set of letters is $\{A, B, C, D, E, F, G, H, I, J\}$,which contains $10$ distinct letters. For $x$,we form words of length $10$ using all $10$ letters without repetition,so $x = 10!$. For $y$,we select $2$ letters to be repeated twice from $10$ available letters in ${}^{10}C_2$ ways. The remaining $6$ positions in the word of length $10$ must be filled by $6$ distinct letters chosen from the remaining $8$ letters,which can be done in ${}^{8}C_6$ ways. The total number of arrangements for these $10$ letters (where $2$ letters appear twice and $6$ letters appear once) is given by the multinomial coefficient $\frac{10!}{2! 	imes 2!}$. Thus,$y = {}^{10}C_2 	imes {}^{8}C_6 	imes \frac{10!}{2! 	imes 2!}$. Calculating the ratio $\frac{y}{x}$: $\frac{y}{x} = \frac{{}^{10}C_2 	imes {}^{8}C_6 	imes \frac{10!}{2! 	imes 2!}}{10!} = \frac{{}^{10}C_2 	imes {}^{8}C_6}{4} = \frac{45 	imes 28}{4} = 45 	imes 7 = 315$.

Words of length $10$ are formed by 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 $y$ be the 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

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