Question

Prove that : \(\large{^n P_r = ^{n-1} P_r + r ^{n-1} P _{r-1} }\)

Answer 1

\(\large{^n P_r = ^{n-1} P_r + r ^{n-1} P _{r-1} }\)

Answer 2

Sorry for that mistake. Please refresh the page to see the edited question or just check the above post by me.

Answer 3

Here is my work :

\( \large{
\textbf{Simplifying LHS to get RHS } \\
^{n-1} P _ r + r ^{n-1} P _ {r-1} = \cfrac{(n-1)!}{(n-1-r)!} + r [ \cfrac{(n-1)!}{(n-r)!} ] \\
\cfrac{(n-1)! (n-r)}{(n-r-1)! (n-r)} + r [ \cfrac{(n-1)!}{(n-r)!} ] \\
\cfrac{(n-1)! (n-r)}{(n-r)!} + r [ \cfrac{(n-1)!}{(n-r)!} ] \\
\cfrac{(n-1)!}{(n-r)!} \{ n-r + r\} \\
\cfrac{(n-1)! n}{(n-r)!} \\
\cfrac{n!}{(n-r)!} = ^n P _r ... \mathsf{Proved!} \\
\\
^{n-1} P _ r + r ^{n-1} P _ {r-1} = \cfrac{n!}{(n-r)!} = ^n P _ r
} \)

WOW I did it :)

Answer 4

:O Well, OS helps in this way too ... Never thought about that.

Answer 5

Well done:)

Answer 6

Thanks @ajprincess ... Actually I was not able to find any way to proceed for the proof. But I did it finally :) btw, you were typing something, was it the solution? Was it the same as mine or something different? If diff. then please let me know.

Answer 7

Actually i was typing part of it so u can do the remaining one. it is the same.:)

Answer 8

Oh k . Thanks for letting me know.

Answer 9

Welcome:)