3u Mathematics Marathon V 1.1 (1 Viewer)

[webby234]

NEXT QUESTION:
this one had me thinking for a while......
Q - five persons play a card game in which one pair play against the other pair. tjhe fifth person acts as referee. in how many ways can this occur?

btw - how do you use a spoiler?

Spoiler

I would say 5 (to choose the ref) * 4C2/2 (number of ways of making two teams of two from four people). So 15

Am I right?

 

[pLuvia]

[spoiler ]*text*[ /spoiler]
But without the spaces

Edit: Meh

 

[hyparzero]

NEXT QUESTION:
this one had me thinking for a while......
Q - five persons play a card game in which one pair play against the other pair. tjhe fifth person acts as referee. in how many ways can this occur?

15

Next Question:
Mr. Jones's bathtub leaks water at a rate that is proportional to the volume of water in the tub. Continuously adding 5L/minute into the tub will just balance the loss through the leak when the tub contains 105 L of water.

One day, Mr. Jones could only allow the water to run for 15minutes before the tub was full. Determine the volume of water initally.

 

[webby234]

NEXT QUESTION (an easy one because I can't think of any ):

Prove that nC1 + nC2...nCn = 2ⁿ

 

[webby234]

Wait a minute - we have different answers - who's right?

Riviet - I think you have to divide by two as selecting AB is equivalent to selecting CD, selecting AC is equivalent to selecting BD etc.

 

[onebytwo]

for the comb. question, apparently the answer is 15, so you guys are right

 

[pLuvia]

Spoiler

So would it be the same if lets say, 6 people were divided up into 3 groups, with groups of 2, so it was 6C2/3?

 

[webby234]

Sorry - just realised I called you Riviet

Spoiler

I think that's a different case because you have to choose three groups. You would choose two from 6 for the first group, then have four left for the second and third groups. So I think you'd have 6C2 * 4C2/3! = 15 The 3! comes in because each different combination can happen in 3! different ways ie if you had AB, CD and EF you could arrange those groups in 3! different ways, but for our purposes they are all equivalent

 

[pLuvia]

Spoiler

Ah ok, I sort of understand it now. So with that previous question, it's 5C1*4C2/2!, because you can arrange the two groups in 2! ways?

 

[webby234]

Yeah - that's my understanding of it anyway.

 

[BIRUNI]

I looked at this solution. this solution is excellent but I have a shorter solution and I thought You might like this:

let n be an even integer. we consider (1+x)^n and (1-x)^n

(1+x)^n=1+nc1 x+nc2 x^2+nc3 x^3+...+ncn x^n
(1-x)^n=1-nc1 x+nc2 x^2-nc3 x^3+...+ncn x^n

(1+x)^n-(1-x)^n=2{nc1 x+ nc3 x^3+...}
let x=1
2^(n-1)=nc1+nc3+...

 

[Riviet]

Riviet - I think you have to divide by two as selecting AB is equivalent to selecting CD, selecting AC is equivalent to selecting BD etc.

And I was like what the...

 

[BIRUNI]

is there any possibility to have geometry proofs by induction in HSC?

 

[Riviet]

is there any possibility to have geometry proofs by induction in HSC?

It's possible (but not likely), you never know what to expect...

 

[BIRUNI]

question: all the letters of the word FACETIOUS are arranged in a line, find the total number of arrangement that contain all the vowels in the order AEIOU but not necessarily together?

 

[onebytwo]

Spoiler

im probably wrong...but 9!/5!

 

[zeek]

Spoiler

arrangement of the vowels first... 5.4.3.2.1=5!
arrangement of the other letters around the vowels = 4!
Total arrangements = 5!4!

is that right

 

[pLuvia]

Spoiler

The vowels can only be in that order so 1 way
4! ways of arranging the other letters
So 4! ways?

 

[onebytwo]

oh dear, 3 different answers!
i didnt give my reasoning, so here it is...

Spoiler

the arrangement AEIOU is one arrangement out of 5!(120) arrangements
the arrangement AEIOU occurs just as many times as the others arrangements like AEUIO, EIUOA, OIUEA etc. so the total number of arrangemnts are 9!(because 9 different letters) and to have the vowels in one particular order we divide by 5!(120)........does that even make sense?......how times did i write "arrangement"?

 

[webby234]

question: all the letters of the word FACETIOUS are arranged in a line, find the total number of arrangement that contain all the vowels in the order AEIOU but not necessarily together?

Spoiler

I think I agree with onebytwo's logic - the probability of that arrangement is 1/5! so 9!/5! appears right.

Alternative method - Set the vowels in order

A E I O U
THen four of the remaining nine spots are left so you can place the consonants in 9P4 ways which is the same as 9!/5! = 3024