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(A) The number of ways of not selecting $(n-r)$ things from $n$ different things is equivalent to selecting $r$ things,which is ${ }^n C_r = { }^n C_{

Vedclass · Accepted Answer

$A$			$B$			$C$			$D$							$V$			$III$			$IV$			$II$

Vedclass · Answer

(A) The number of ways of not selecting $(n-r)$ things from $n$ different things is equivalent to selecting $r$ things,which is ${ }^n C_r = { }^n C_{n-r}$. Thus,$(A) \rightarrow (V)$.$(B)$ We have $(n-r+1) \cdot { }^n C_{r-1} = (n-r+1) \cdot \frac{n!}{(r-1)!(n-r+1)!} = \frac{n!}{r!(n-r)!} \cdot r = r \cdot { }^n C_r$. Thus,$(B) \rightarrow (III)$.$(C)$ The number of ways of selecting at least $(n-r)$ things from $n$ different things is ${ }^n C_{n-r} + { }^n C_{n-r+1} + \ldots + { }^n C_n$. This is equal to $2^n - ({ }^n C_0 + { }^n C_1 + \ldots + { }^n C_{n-r-1})$. Since ${ }^n C_k = { }^n C_{n-k}$,this matches the expression $2^n - 1 - n - { }^n C_2 - \ldots - { }^n C_r$ (assuming $r$ is small). Thus,$(C) \rightarrow (IV)$.$(D)$ $(n-r)({ }^{n-1} C_{r-1} + { }^{n-1} C_r) = (n-r)({ }^n C_r) = (n-r) \cdot \frac{n!}{r!(n-r)!} = \frac{n!}{r!(n-r-1)!} = (r+1) \cdot \frac{n!}{(r+1)!(n-r-1)!} = (r+1) \cdot { }^n C_{r+1}$. Thus,$(D) \rightarrow (II)$.

	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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