This is METH-ness! (H/w help trend).

[the.watchman]

When it gets closer to exam time, it may be worthwhile you trying to answer all the MM questions on VN, they can be good practice questions as well (that's what I'm doing! )

Although you're a good year ahead. :p

Excuse me, brightsky! Who's the maths freak?

[Rockim]

I need help with binomial expansion

Find the coefficient of x^3 in the expansion of:

(x-1/x)^10

[appianway]

In terms of cheatsheets, they shouldn't be needed. They're a good back up, but you probably won't end up using them. I didn't use mine for physics, and I definitely wasn't alone... most people didn't!

[the.watchman]

There is no term, I think, so the co-efficient of is 0.

This is because there is no (between 0 and 10) such that becomes

[Rockim]

alright thanks watchman

[brightsky]

I need help with binomial expansion

Find the coefficient of x^3 in the expansion of:

(x-1/x)^10

Yep no



The possible "bases" are

As you can see, when we have a even power of , the power of takes an even number away (or zero) (e.g. . If we have an odd power of , the power of , takes an odd number away (e.g. . This is to say, no matter if we have an odd or even power of , the resulting "base" will always have an even power, or a "negative" even power (like -2). And because has an odd power, it will never exist...

Dodgy explanation I know...:p

[Rockim]

What about this one for arrangements?

Find the number of ways the letters in the word ROOSTER can be arranged if each letter must be used and:

the R's must be together and the O's must be together.

All my classmates, including myself, have shown the working out differently. Can someone please clarify? the answer is 120

[the.watchman]

alright thanks watchman

No prob.

brightsky, a funny way to prove it would be by using simultaneous equations:

and , with only integer solutions and

[superflya]

7 letters and both O's and R's must be together.... ??

[andy456]

What about this one for arrangements?

Find the number of ways the letters in the word ROOSTER can be arranged if each letter must be used and:

the R's must be together and the O's must be together.

All my classmates, including myself, have shown the working out differently. Can someone please clarify? the answer is 120

5! = 5x4x3x2x1
    = 120

[the.watchman]

What about this one for arrangements?

Find the number of ways the letters in the word ROOSTER can be arranged if each letter must be used and:

the R's must be together and the O's must be together.

All my classmates, including myself, have shown the working out differently. Can someone please clarify? the answer is 120

Two groups: [R,R] & [O,O], plus S,T,E anywhere

So there are, sort of, six available positions (5!=120 possibilities)

I'm not so sure though...

[kyzoo]

When it gets closer to exam time, it may be worthwhile you trying to answer all the MM questions on VN, they can be good practice questions as well.

=/ Sounds very inefficient: The time you take in finding questions, and the fact (at least for me) that it's easy to be distracted on the Internet.

[the.watchman]

When it gets closer to exam time, it may be worthwhile you trying to answer all the MM questions on VN, they can be good practice questions as well.

=/ Sounds very inefficient: The time you take in finding questions, and the fact (at least for me) that it's easy to be distracted on the Internet.

Yeah, I guess, but I multitask (answer own questions & surf internet/VN)

[Rockim]

i need help with this combination question:

5!/r!(5-r)!=10 or (5Cr=10)

find the pronumeral 'r'

sorry i still don't know how to show my maths questions in the right format

[the.watchman]

You can use Pascal's triangle (a bit cheap, I know), it's quicker than algebraically (answer is 2,3)