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 remaining

Vedclass · Accepted Answer

$21x = 4y$

Vedclass · Answer

(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,$21x = 4y$.

	List-$I$		List-$II$
$(A)$	The number of ways of not selecting $(n-r)$ things from $n$ different things	$(I)$	$1+n+{ }^n C_2+\ldots+{ }^n C_r$
$(B)$	$(n-r+1) \cdot{ }^n C_{r-1}$	$(II)$	$(r+1) \cdot{ }^n C_{r+1}$
$(C)$	The number of ways of selecting at least $(n-r)$ things from $n$ different things	$(III)$	$r\left({ }^n C_r\right)$
$(D)$	$(n-r)\left({ }^{n-1} C_{r-1}+{ }^{n-1} C_r\right)$	$(IV)$	$2^n-1-n-{ }^n C_2-\ldots-{ }^n C_r$
		$(V)$	${ }^n C_{n-r}$

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