Question

find the value of n or r
4np3=5(n-1)p3​

Answer 1

Step-by-step explanation:

Given

5.

n

P

3

=4.

n+1

P

3

3n(n−1)(n−2)=4(n+1)n(n−1)

5(n−2)=

n(n−1)

4(n+1)n(n−1)

5(n−2)=4(n+1)

5n−10=4n+4

5n−4n=4+10

n=14

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Answer 2

$\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf4 \times {^{n}P_3 = 5 \times ^{n - 1}P_3} \end{cases}\end{gathered}\end{gathered}$

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

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

We know that,

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

Consider,

$\rm :\longmapsto\:4 \times {^{n}P_3 = 5 \times ^{n - 1}P_3}$

$\rm :\longmapsto\:4 \times \dfrac{n!}{(n - 3)!} = 5 \times \dfrac{(n - 1)!}{(n - 1 - 3)!}$

$\rm :\longmapsto\:\dfrac{4n(n - 1)!}{(n - 3)(n - 4)!} = \dfrac{5(n - 1)!}{(n - 4)!}$

$\rm :\longmapsto\:\dfrac{4n}{n - 3} = 5$

$\rm :\longmapsto\:4n = 5n - 15$

$\bf\implies \:n \: = \: 15$

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Additional Information :-

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

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

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