Binomial help (1 Viewer)

[tehsky]

Can anyone show me the full expansion..? as in what cancles with what factorial etc..the addition bit is confusng me :S

 

[vulgarfraction]

1- draw out a corresponding pascal's triangle. link- http://mathworld.wolfram.com/PascalsTriangle.html

2 (arrows bit)- note that a! = a* (a-1)!, and put in a common denominator.. Stuff cancels.

 

[tehsky]

yeah i know stuff cancels..
but i just want to see how and what cancels out; not automatically know it cancels =/

 

[wrxsti]

mmhmm i had this problem... but i fink i understand how they cancelled out...
you know the (n - k +1) ! i fink dey broke it up into (n -k)! x (n-k+1)
they then equated demoniators..... blah blah blah... and cancelled out..
im pretty sure.... thats what happened

 

[Buiboi]

Can anyone show me the full expansion..? as in what cancles with what factorial etc..the addition bit is confusng me :S

you understand how they got the denominator right?? (its just common denominator crap)

but heres some data:
r! = r (r-1)(r-2)(r-3)....= r (r-1)! hence, r!>(r-1)! by r

so basically of the r!, the (r-1)(r-2)... all cancel out which is = to (r-1)! thats why yo ucancel them, the r! is larger by r...you get me?

(n-r+1)!= (n-r+1) (n-r) (n-r-1)....= (n-r+1) (n-r)! hence (n-r+1)! >(n-r)! by (n-r+1)

same thing as i said b4?

oh shit i dunno how to explain lol.....but like can you work it out from what ive shown?

 

[Trebla]

From: n! / [r!.(n - r)!] + n! / [(r - 1)!.(n - r + 1)!]

Break the r! into r.(r - 1)! and the (n - r + 1)! into (n - r + 1).(n - r)!
= n! / [r.(r - 1)!.(n - r)!] + n! / [(r - 1)!.(n - r + 1).(n - r)!]

To make life easier, factorise out: n! / [(r - 1)!.(n - r)!]
= n! / [(r - 1)!.(n - r)!] { 1 / r + 1 / (n - r + 1) }

Now for the expression in { }, put over a common denominator:
= n! / [(r - 1)!.(n - r)!] x { [(n - r + 1) + r] / r.(n - r + 1) }

Now put the factor back in:
= [n!.(n - r + 1) + r.n!] / [r.(r - 1)!.(n - r + 1)(n - r)!]
= [n!.(n - r + 1) + r.n!] / [r!.(n - r + 1)!]

 

[tehsky]

From: n! / [r!.(n - r)!] + n! / [(r - 1)!.(n - r + 1)!]

Break the r! into r.(r - 1)! and the (n - r + 1)! into (n - r + 1).(n - r)!
= n! / [r.(r - 1)!.(n - r)!] + n! / [(r - 1)!.(n - r + 1).(n - r)!]

To make life easier, factorise out: n! / [(r - 1)!.(n - r)!]
= n! / [(r - 1)!.(n - r)!] { 1 / r + 1 / (n - r + 1) }

Now for the expression in { }, put over a common denominator:
= n! / [(r - 1)!.(n - r)!] x { [(n - r + 1) + r] / r.(n - r + 1) }

Now put the factor back in:
= [n!.(n - r + 1) + r.n!] / [r.(r - 1)!.(n - r + 1)(n - r)!]
= [n!.(n - r + 1) + r.n!] / [r!.(n - r + 1)!]

ahhhh

you sir are a genii
thank you so much