Words of length 10 are formed using the lette | vedclass

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(A) The total number of letters available is $10$. For $x$,we form words of length $10$ with no repetition,which is a permutation of $10$ distinct let

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The total number of letters available is $10$. For $x$,we form words of length $10$ with no repetition,which is a permutation of $10$ distinct letters: $x = 10!$. For $y$,we choose $1$ letter to be repeated twice from $10$ letters in $^{10}C_1$ ways. We then choose $8$ other letters from the remaining $9$ letters in $^9C_8$ ways. The total number of arrangements of these $10$ letters (where one is repeated twice) is $\frac{10!}{2!}$. Thus,$y = {}^{10}C_1 	imes {}^{9}C_8 	imes \frac{10!}{2!}$. Calculating the ratio: $\frac{y}{9x} = \frac{{}^{10}C_1 	imes {}^{9}C_8 	imes \frac{10!}{2!}}{9 	imes 10!} = \frac{10 	imes 9}{9 	imes 2} = 5$.

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

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