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Permutations algebra

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Reply 20
Avatar for SubAtomic
SubAtomic
OP
11
Original post by notnek
=n!(k+1)(n−(k+1))(n−(k+2))⋯1×k!=n!(k+1)(n−(k+1))!×k!\displaystyle = \frac{n!}{(k+1)(n-(k+1))(n-(k+2))\cdots 1 \times k!} = \frac{n!}{(k+1)(n-(k+1))! \times k!}


Now notice that (k+1)×k!(k+1) \times k! is equal to (k+1)!(k+1)! in the same way that e.g. 5x4!=5! or 8x7!=8!.

Does it make sense now?


So does it become

n!(k+1)!(n−(k+1))!\displaystyle\frac{n!}{(k+1)!(n-(k+1))!}
Reply 21
Avatar for SubAtomic
SubAtomic
OP
11
Original post by notnek
I got confused because you didn't write down the nCk=\displaystyle^n C_k = part when posting eqn (5.6) although I probably should've noticed that they were equal.


I may be a bit mystical with my questions at times due to not doing the 'norm' A level type stuff.


Original post by notnek


nCk+1=n(n−1)⋯(n−k))(k+1)!\displaystyle^n C_{k+1} = \frac{n(n-1)\cdots (n-k))}{(k+1)!}

and you can write this in a different way:

=n(n−1)⋯(n−(k−1))(n−k)k!×(k+1)=n(n−1)⋯(n−(k−1))k!×n−kk+1=\displaystyle \frac{n(n-1)\cdots (n-(k-1))(n-k)}{k!\times (k+1)} = \frac{n(n-1)\cdots (n-(k-1))}{k!} \times \frac{n-k}{k+1}



Think this is what I was looking for, thanks:cool:
Reply 22
Avatar for Notnek
Notnek
21
Original post by SubAtomic
So does it become

n!(k+1)!(n−(k+1))!\displaystyle\frac{n!}{(k+1)!(n-(k+1))!}

Yes and by definition this is equal to nCk+1\displaystyle^n C_{k+1}
Reply 23
Avatar for SubAtomic
SubAtomic
OP
11
Original post by notnek
Yes and by definition this is equal to nCk+1\displaystyle^n C_{k+1}


Much appreciated. Will have to expand all this a few times to get it.
(edited 11 years ago)

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