Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let x be | vedclass

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(C) The set $S$ contains $9$ distinct digits. For $x$: We choose $1$ digit to be repeated twice, which can be done in ${}^{9}C_{1}$ ways. The remainin

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$21x = 4y$

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(C) The set $S$ contains $9$ distinct digits. For $x$: We choose $1$ digit to be repeated twice, which can be done in ${}^{9}C_{1}$ ways. The remaining $7$ digits are chosen from the remaining $8$ digits in ${}^{8}C_{7}$ ways. The total number of arrangements is $\frac{9!}{2!}$. Thus, $x = {}^{9}C_{1} 	imes {}^{8}C_{7} 	imes \frac{9!}{2!} = 9 	imes 8 	imes \frac{9!}{2} = 36 	imes 9!$. For $y$: We choose $2$ digits to be repeated twice, which can be done in ${}^{9}C_{2}$ ways. The remaining $5$ digits are chosen from the remaining $7$ digits in ${}^{7}C_{5}$ ways. The total number of arrangements is $\frac{9!}{2! 	imes 2!}$. Thus, $y = {}^{9}C_{2} 	imes {}^{7}C_{5} 	imes \frac{9!}{2! 	imes 2!} = \frac{9 	imes 8}{2} 	imes \frac{7 	imes 6}{2} 	imes \frac{9!}{4} = 36 	imes 21 	imes \frac{9!}{4} = 189 	imes 9!$. Calculating the ratio: $\frac{x}{y} = \frac{36 	imes 9!}{189 	imes 9!} = \frac{36}{189} = \frac{4}{21}$. Therefore, $21x = 4y$.

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