Extension One Revising Game (1 Viewer)

[lyounamu]

i've asked a few people and the answers i get range from 100k-500k....but now even less lol

What's the point of asking if you don't have the answer?

 

[gurmies]

Agreed^

Although when lolokay comes, he'll get the answer indefinitely

 

[cyl123]

do mine please

All the letters of the word ENCUMBRANCE are arranged in a line. Find the total number of arrangements, which contain all the vowels in alphabetical order but separated by at least one consonant.

This is the technique used for questions involve placing 2 different sets of objects such that one set of objects must be separate from each other (such as placing 5 girls and 3 boys with all boys separate)

So first of all place all the consanents (N C M B R N C) in a line. There are in total 7!/4 arrangements.

Now consider the spaces in between each consanent as a possible position to place a vowel, since in this way the vowels will be separated. There are 8 spaces in total (counting space outside each end. The vowels are A E U E.

So now only consider number of arrangement where vowels are separate not necessarily in alphabetical order. There are 4 letters to place so there are 8 spaces to place first letter, 7 spaces to place second letter and so on. There are 8P4/2 ways to do this (counting repetition)

Now note that there are 12 possible ways to arrange the letters A U E E in a straight line but only one arrangement is in alphabetical order. So when the letters are placed in the spaces, when ignoring the consonants, only 1 in 12 arrangements are in alphabetical order. Thus out of the 8P4/2 arrangements, only 1 in 12 are in alphabetical order.

Thus total number of arrangement = 7!/4*8P4/2*1/12 = 88200 arrangements.

i think this is right...

 

[Timothy.Siu]

yeah it is, thanks for that

 

[lolokay]

Now consider the spaces in between each consanent as a possible position to place a vowel, since in this way the vowels will be separated. There are 8 spaces in total (counting space outside each end. The vowels are A E U E.

So now only consider number of arrangement where vowels are separate not necessarily in alphabetical order. There are 4 letters to place so there are 8 spaces to place first letter, 7 spaces to place second letter and so on. There are 8P4/2 ways to do this (counting repetition)

Now note that there are 12 possible ways to arrange the letters A U E E in a straight line but only one arrangement is in alphabetical order. So when the letters are placed in the spaces, when ignoring the consonants, only 1 in 12 arrangements are in alphabetical order. Thus out of the 8P4/2 arrangements, only 1 in 12 are in alphabetical order.

Thus total number of arrangement = 7!/4*8P4/2*1/12 = 88200 arrangements.

i think this is right...

to order the vowels, rather than treating them as different letters I would just straight away consider them as identitcal (since they can't be ordered), so to order them it is 8C4 rather than 8P4/2 (ie. it is the number of ways of choosing 4 of the 8 spots to have a vowel in them). this simplifies it a bit (still the same answer)

 

[addikaye03]

Question: ( to keep things going, its 4U tho)

a)If z1=1+i,z2=rt3-i, find the moduli and principal arguments of z1,z2 and z1/z2
b) If z=(1+i)/(rt3-i), find the smallest postive integer n such that z^n is real and evaluate z^n for this integer n.

 

[Trebla]

Here's a random one

Prove that log₂3 is irrational.

 

[addikaye03]

Here's a random one

Prove that log₂3 is irrational.

nup, i cant get a lead. Im sure its something small. Something to do with change of base maybe.

 

[Aerath]

ln 3 / ln 2?

Since both are irrational? I have no idea.

 

[lyounamu]

Here's a random one

Prove that log₂3 is irrational.

log(2) 3 = ln(3)/ln(2)

hm....

wait...

ln(3)/ln(2) = (ln(2) + ln(1.5))/ln(2) = 1 + ln(1.5)/ln(2)

Ah...doesn't make much difference...

 

[Pwnage101]

Question: ( to keep things going, its 4U tho)

a)If z1=1+i,z2=rt3-i, find the moduli and principal arguments of z1,z2 and z1/z2
b) If z=(1+i)/(rt3-i), find the smallest postive integer n such that z^n is real and evaluate z^n for this integer n.

isnt that MX2...

 

[azureus88]

Suppose log(2) 3 is rational

then log(2)3=p/q
2^(p/q) = 3
2^p = 3^q

but LHS is even and RHS is odd so by contradiction, log(2)3 is irrational.

 

[lyounamu]

Suppose log(2) 3 is rational

then log(2)3=p/q
2^(p/q) = 3
2^p = 3^q

but LHS is even and RHS is odd so by contradiction, log(2)3 is irrational.

so you are basically saying that p and q are irrational from there...

 

[Pwnage101]

Suppose log(2) 3 is rational

then log(2)3=p/q
2^(p/q) = 3
2^p = 3^q

but LHS is even and RHS is odd so by contradiction, log(2)3 is irrational.

yeh, i think that's a great way to prove something is irrational - by contradiction with the p/q stuff

well done

(i would also add that p and q must be integers, q cannot = 0, and p & q have no common factors, if you wanted an even 'better' solution)

 

[Pwnage101]

so you are basically saying that p and q are irrational from there...

yeh he is, he missed a crutial step for ppl who havent seen a proof like this to understand it:

the definition of a rational number is a number that can be expressed in the form of p/q, where p & q are integers and q cannot = 0

so he assumes this, but if assume this u come up with a contradiction, which means it cannot be true, ie the number is not rational

 

[lyounamu]

yeh, i think that's a great way to prove something is irrational - by contradiction with the p/q stuff

well done

(i would also add that p and q must be integers, q cannot = 0, and p & q have no common factors, if you wanted an even 'better' solution)

Yes, that's very interesting. Answering using the notion of contradiciton seems very useful. I guess I will take that method on board, haha.

 

[Trebla]

There are proofs that e and π are irrational as well using proof by contradication, but they are far more complicated than this lol (see Extension 2 HSC 2001 and 2003)

 

[Pwnage101]

There are proofs that e and π are irrational as well using proof by contradication, but they are far more complicated than this lol (see Extension 2 HSC 2001 and 2003)

lol i still remember my first lesson of 3 unit maths in year 11...i had a great teacher, and the very first lesson he gave us all these definitions of rational numbers, etc then showed us the proof that e is irrational via that contradiction method

i still have my book, and although i didnt understand it at the time, it's times like these when you see that it has actually been tested in MX2 (with direction, of course) that you appreciate a great teacher

 

[addikaye03]

Any more Q Treb?

 

[gurmies]

Pwnage, on the first day of year 11 3 unit maths, our teacher did the exact same thing!