questions - permutations, GAD (1 Viewer)

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pamplemousse
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I don't know what is wrong with me, I seemed to have done too much 2 u revision and not enough 3 u. Can someone help me with these questions? Thank you and sorry for being such a dumbass. :( I really shouldn't be doing ext 1, I'm a humanities and english girl.

How many arrangements of the word barbados can occur in a straight line if the 2 bs cannot be adjacent?

When the graph of dy/dx has a stationary pt, min, max etc what happens to the corresponding function f(x)?

GAH! I should know this.
 

wogboy

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How many arrangements of the word barbados can occur in a straight line if the 2 bs cannot be adjacent?
Without the restriction of the 2 b's together, the no of unique words formed is:
8! / (2! * 2!)

Of these unique words, 7 * (6! / 2!) are not allowed (two b's occur sequentially)

So the answer is 8! / (2! * 2!) - 7 * (6! / 2!) = 7560
 

Sharky

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cherryblossom said:
When the graph of dy/dx has a stationary pt, min, max etc what happens to the corresponding function f(x)?
When the graph of dy/dx has either a min or a max turning point, the corresponding graph of f(x) will have a point of inflection.

And if dy/dx has a point of inflection, ... that one's a bit harder to visualize, i'll see what I can do
 

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ty, I'd really appreciate it. :)

Lol, I really am disorganised - does anyone know where I can get any resources on proofs using the binomial theorem?
 

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lol, I'm not blonde, it's just I have a really crappy text without anything on binomials in it.
 

Rorix

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i think she (i think its a she at least?) is referring to the proofs involving nCr and the expansion of (1+x)^n

I don't really know what you mean by resources, try looking at past papers with binomail qs and worked solns?
 

Rorix

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oh i only read the posts i write :cool:
 

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Would a guy call himself cherryblossom?

No, no kindly refrain from answering that question, I've just decided I don't want to know.

What I meant was...
Proofs using the binomial theorem, using integration, subs, differentiation and what not - any sample questions with solutions will do
 

Rorix

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oh I don't have any at hand but i could probably make some up tomorrow >_>

oh, and guys have called themself worse than cherryblossom ;)
 

Sharky

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Possible proof question...

Prove that:

nCr = (n-1)C(r-1) + (n-1)Cr

(This sorta makes sense if u look at pascal's triangle, it's just that any one equals the sum of the two above it!)

RHS = (n-1)C(r-1) + (n-1)Cr
= (n-1)!/[(n-1-(r-1))!*(r-1)!] + (n-1)!/[(n-r-1)!*r!]
= (n-1)!/[(n-r)!*(r-1)!] + (n-1)!/[(n-r-1)!*r!]
= (n-1)!/[(n-r)*(n-r-1)!*(r-1)!] + (n-1)!/[(n-r-1)!*r*(r-1)!]
= [(n-1)!*r+(n-1)!*(n-r)]/[(n-r)*(n-r-1)!*r*(r-1)!]
= [(n-1)!*(r+n-r)]/[(n-r)!*r!]
= [(n-1)!*n]/[(n-r)!*r!]
= n!/[(n-r)!*r!]
= nCr
= LHS

Sorry as that probably looks messy. but i tried to expand the denominator in there to make it easier to follow, it wasn't actually necessary.

I hope that helps, rather than confuses you. :)

If you REALLY can't actually understand this one at all, then just reply and i'll write it up into a word document for you to read more easily.
 

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Yeah I follow, that's about the ONLY proof I have in my text lol...

Thank you, do you have any involving, um, for example, (x + 1)*n(x + 1)*n = (x + 1)*n?
 

ngai

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cherryblossom said:
(x + 1)*n(x + 1)*n = (x + 1)*n
...

are those * meant to be ^
and the n meant to be 2n on the right?
and is it meant to be stuff like by comparing coefficients of x^r,
show that 2nCr = nC0*nCr + nC1*nC(r-1) + ... + nCr*nC0?
 

Rorix

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there's many proofs you can make your own:)

e.g.
by considering the identity (1+x)^8 (1+x)^8 = (1+x)^16
prove (8c0)^2 + (8c1)^2 +......+(8c8)^2 = 16c8
 

paper cup

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yes yes that's what I meant.

Anyone have any examples of proofs though?
 

Rorix

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well for my above question
expanding LHS gives (8c0 + 8c1 x + 8c2 x^2 + .....8c8 x^8)(8c0 + 8c1 x + 8c2 x^2 + .....8c8 x^8)
expaning RHS gives (16c0 + 16c1x+16c2x^2+..........+16c16 x^16)

Comparing coefficents of x^8
on LHS: 8c0 8c8 + 8c1 8c7 + 8c2 8c6 +..........+8c8 8c0
on RHS: 16c8
therefore 8c0 8c8 + 8c1 8c7 +.......+8c8 8c0 = 16c8
but 8cr = 8c(8-r)
so (8c0)^2 + (8c1)^2 + (8c2)^2 +.......+ (8c8)^2 = 16c8
 

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