The discussion revolves around proving the equation \((n+1)nPr=(n+1)P(r+1)\), which pertains to permutations in combinatorial mathematics.
Participants are engaged in examining their approaches to the proof, with some clarifying the distinction between different forms of permutations. There is an ongoing exploration of how to manipulate the left-hand side to match the right-hand side of the equation.
Some participants express uncertainty about the manipulation of factorial expressions and the implications of their substitutions. There is a focus on ensuring that equalities are maintained throughout the proof process.
I don't think (n + 1) * n! = [n * (n + 1)]!Pual Black said:Homework Statement
Hello
I have to proof this
##(n+1)nPr=(n+1)P(r+1)##
The attempt at a solution
if i substitute this in
##nPr=\frac{n!}{(n-r)!}##
I get this
##(n+1)nPr=\frac{[(n+1)n]!}{[(n+1)n-r]!}##
This will give
##(n+1)nPr=\frac{(n^2+n)!}{(n^2+n-r)!}##
Now i don't know how to continue.
Yes. It's (n+1)(nPr).Pual Black said:Did you mean that (n+1)nPr isn't the same as ((n+1)n)Pr ??
Yes. And since you are dealing with equalities, not inequalities, it doesn't matter which side you start with to arrive at the other.Pual Black said:Ok thanks. Now let's go on
##(n+1)nPr=(n+1)\frac{n!}{(n-r)!}##
I have to get this
##\frac{(n+1)!}{[(n+1)-(r+1)]!}##
to proof the R.H.S
But if i start with the Right side
##(n+1)P(r+1)=\frac{(n+1)!}{[(n+1)-(r+1)]!}=(n+1)\frac{n!}{(n-r)!}=(n+1)nPr=L.H.S##
But how to start with left side and proof the right side ?
I can -1 +1 to the denominator to get the (n+1)-(r+1)
But can i say that (n+1)n!=(n+1)! ?
I think yes of course
Ok. Thank you very much for your help.haruspex said:Yes. And since you are dealing with equalities, not inequalities, it doesn't matter which side you start with to arrive at the other.