mmmmmoooooorrrrrrreeeeeeeeeee (1 Viewer)

[spice girl]

Originally posted by BlackJack
Anyway, spice girl I have a question, just as an aside.
Have you competed in the UNSW schools maths comp (if you've heard of it)? Or the maths olympiads?

Yes, both, but not internationally, not good enough...

 

[Milly]

Originally posted by p00_p00
Well if its FREQUENT

shouldnt it still be 8!/2! ??????

It can't just be 8!/2! because you only divide by 2! when there's 2 e's. So you have to treat the combinations with 2 e's and the combinations with less than 2 e's separately. I think. :worried:

 

[p00_p00]

??????


I dont get

there are only 2 reoccurring letters. E

There are a total of 8 letters and 3 too chose from

Hence 8P3/2!

 

[spice girl]

p00_p00,

It's not a question of wot's wrong with your answer, it's a question of wot's wrong with ours?

If you simplify a solution and get a different answer, u should at least suspect that you've simplified it too much.

I think you should do a bit of brushing up on your probability...if you do 3u or higher

 

[p00_p00]

wats wrong with urs???


well if were talkin about words, with 8 letters, isnt each letter (besides the E) individuals and unique? Hence order must be important. First mistake with urs was u used C. It should b P

So it should be 8P3/2! (no. of letters that repeat)

 

[Milly]

Originally posted by p00_p00
wats wrong with urs???


well if were talkin about words, with 8 letters, isnt each letter (besides the E) individuals and unique? Hence order must be important. First mistake with urs was u used C. It should b P

So it should be 8P3/2! (no. of letters that repeat)

She did use C, but that was just to get all the possible selections. Then she multiplied it by 3! to take into account the different arrangements (which is like using P, but taking into account the selections with 2 e's and the selections without 2 e's).

(Because you only have to divide by 2! if there's 2 e's. So you derive the number of selections separately, and then find the no. of arrangements differently in each case.)

 

[spice girl]

Hehehe...i dun know why I'm bothering, but anyway, for the spirit of generosity:

Who's the teacher who taught you to separate every question into either "order does matter" or "order doesn't matter"?

When we have repeat letters (the E), but only picking some (and not all) letters, we have to apply a hybrid between the two - order matters "sometimes".

It's obvious that out of all the words you could make, you only get double E's in some 3-letter words, but not in all.

When we get a partial situation like this, we've got to separate them into cases: repeat letters or no-repeat letters - and no-repeat letters require P (7*6*5) - because there are only 7 different letters, not 8.

Repeat letters - u can figure it out by writing them all out - i'll tell you there's 18.

Add the two together.
***end***

 

[p00_p00]

i dont get

ill just phone up advice line instead

 

[BlackJack]

Originally posted by spice girl

Yes, both, but not internationally, not good enough...

I thought you did both... definitely good enough. Same story here (well actually, didn't even make selection school in the olympiads).

 

[spice girl]

haha...i got beaten in both 3u and 4u by this friend i helped in maths in the past...similar to u, blakjak

 

[underthesun]

i say any methods that gets 318 works. And it just makes so much logic in my head, the other answers are wrong

 

[spice girl]

Originally posted by underthesun
i say any methods that gets 318 works. And it just makes so much logic in my head, the other answers are wrong

How about this method?

We have to count the number of "3" letter words from a word with "1" repeated letter and "8" letters in total.

Hence the answer is "3-1-8" = 318



This method makes so much logic in my head.

 

[Minai]

lol
I wonder if thats applicable to any permuatation situation...!

 

[underthesun]

Originally posted by spice girl


How about this method?

We have to count the number of "3" letter words from a word with "1" repeated letter and "8" letters in total.

Hence the answer is "3-1-8" = 318



This method makes so much logic in my head.

Brillient!!!!

you should run for the nobel prize award

 

[-=«MÄLÅÇhïtÊ»=-]

poo poo, i bumped into a ques like this b4 and i used the method u suggested, the answer was exactly the same. But my teacher said it's not how u're meant to do it so i don't use it. I 4got wat the reason was, but i fink its coz the 2 Es will be used more den once and 2! ain't gonna do the trick.

I've never done a ques like this, the only one close with wen u have say 7 letters and u make a 6 letter word. Dat's alot easier coz u cna omit 1 letter at a time, den add them all up. It's slow, but it's how we were taught.

Spice gal i like ur method, i cant find anything wrong with it.

But the earlier suggested answer was 318?
I think this may be why:
8p3, no restrictions, imagine if there weren't 2 Es and the 2nd E was a Z or sumfin
You would have 336 combos

The words that are repeated are the ones that contain 2 Es, so as spicegal worked out earlier, that's 18. So there are 18 words that are repeated out of the 336 possible combos

Therefore no. of unique combos:
336 - 18 = 318

I dunno if this method is legit. But it explains how they got their answer.

 

[BlackJack]

Yeah, I said that on the first page... also said there was a problem with picking the E's just once...

 

[spice girl]

This question's probably not 3u, but anyway, for your entertainment:

You have an infinite number of apples, oranges, pears, and watermelon.

How many different ways are there in picking 11 pieces of fruit {edit: to put into a basket}?

 

[underthesun]

is that a super binomial question?

one way to do it is

(a + b + c + d)^11

find the total of all the coefficients, but i can't do that

 

[McLake]

Originally posted by spice girl
This question's probably not 3u, but anyway, for your entertainment:

You have an infinite number of apples, oranges, pears, and watermelon.

How many different ways are there in picking 11 pieces of fruit?

4^11 ?

 

[spice girl]

Originally posted by McLake
4^11 ?

remember order doesn't matter, it's a basket of fruit - i forgot to mention this...