Question

(1)
Find the value of "n" or "r":
np3: (n+1)P3=3:4​

Answer 1

$\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf{^{n}P_3 : \: ^{n + 1}P_3 = 3 : 4} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{n }\end{cases}\end{gathered}\end{gathered}$

$\large\underline{\sf{Solution-}}$

We know that,

$\boxed{\sf \:^{n}P_r=\dfrac{n!}{(n-r)!}} \:$

Given that

$\rm :\longmapsto\:^{n}P_3 :^{n + 1}P_3 = 3 : 4$

$\rm :\longmapsto\:\dfrac{n!}{(n - 3)!} : \dfrac{(n + 1)!}{(n + 1 - 3)!} = 3 :4$

$\rm :\longmapsto\:\dfrac{n!}{(n - 3)!} : \dfrac{(n + 1)!}{(n - 2)!} = 3 :4$

$\rm :\longmapsto\:\dfrac{n!}{(n - 3)!} : \dfrac{(n + 1)n!}{(n - 2)(n - 3)!} = 3 :4$

$\rm :\implies\:\dfrac{n - 2}{n + 1} = \dfrac{3}{4}$

$\rm :\longmapsto\:4n - 8 = 3n + 3$

$\bf\implies \:n \: = \: 11$

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Additional Information :-

Each of the different arrangement which can be made by taking some or all of a number of things is called a permutation. 

$\boxed{ \sf \: ^{n}P_0 = 1}$

$\boxed{ \sf \: ^{n}P_n = n!}$

$\boxed{ \sf \: ^{n}P_r = n\bigg( \: ^{n - 1}P_{r - 1}\bigg)}$