If ^n{P_r} = 840 and ^n{C_r} = 35,then n is e | vedclass

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(D) We know that $^n{P_r} = ^n{C_r} 	imes r!$. Substituting the given values: $840 = 35 	imes r!$. $r! = \frac{840}{35} = 24$. Since $4! = 24$,we ha

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$7$

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(D) We know that $^n{P_r} = ^n{C_r} 	imes r!$. Substituting the given values: $840 = 35 	imes r!$. $r! = \frac{840}{35} = 24$. Since $4! = 24$,we have $r = 4$. Now,using the formula $^n{C_r} = 35$ with $r = 4$: $^n{C_4} = \frac{n(n-1)(n-2)(n-3)}{4 	imes 3 	imes 2 	imes 1} = 35$. $n(n-1)(n-2)(n-3) = 35 	imes 24 = 840$. Testing values,for $n = 7$: $7 	imes 6 	imes 5 	imes 4 = 840$. Thus,$n = 7$.

If $^n{P_r} = 840$ and $^n{C_r} = 35$,then $n$ is equal to

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