(C) Given the equation: $^nP_4 = 30 \times ^nC_5$Using the formulas $^nP_r = \frac{n!}{(n-r)!}$ and $^nC_r = \frac{n!}{r!(n-r)!}$,we have:$\frac{n!}{(

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(C) Given the equation: $^nP_4 = 30 \times ^nC_5$Using the formulas $^nP_r = \frac{n!}{(n-r)!}$ and $^nC_r = \frac{n!}{r!(n-r)!}$,we have:$\frac{n!}{(n-4)!} = 30 \times \frac{n!}{5!(n-5)!}$Divide both sides by $n!$:$\frac{1}{(n-4)(n-5)!} = \frac{30}{120(n-5)!}$$\frac{1}{n-4} = \frac{30}{120} = \frac{1}{4}$$n - 4 = 4$$n = 8$

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

If $^nP_4 = 30 \times ^nC_5$,then $n = \dots$

A
$6$
B
$7$
C
$8$
D
$9$

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If $^nP_4 = 30 \times ^nC_5$,then $n = \dots$

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$I.$ $Zn +$ conc. $HNO_3$ (hot) $\rightarrow Zn(NO_3)_2 + X + H_2O$
$II.$ $Zn +$ dil. $HNO_3$ (cold) $\rightarrow Zn(NO_3)_2 + Y + H_2O$
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