Maths quiz

[ainwood]

10.

Assuming that none of the inerviewees lied.

 

[Zwelgje]

Hmmmm, you didn't fall for the trick. Many people will start calculating when they see the word statistics but the answer is really simple: 10
Ainwood, your turn.

 

[ainwood]

Right.

On the attached diagram, you have to go from "A" to "B". However, at each intersection, you have to go down or to the right ONLY.

How many different routes are there?

 

[ainwood]

Come on guys! Its not that hard!

 

[Lucky]

Of course it´s not. But IF I understood you rules correctly it would simply be a matter of counting. The problem is that this reduces your question to simple hard work, not thinking. There are plenty of ways in your diagramm, more than a lot of us are willing to count.

 

[ainwood]

There's a MUCH easier way than just plotting through the different routes, and then counting them.

I'll post it later (damn work getting in the way...)

 

[AVN]

So, there is a much easier way to solve this problem.
(I also thought I had to count all the possibilities).

In that case you will teach us something valuable.

 

[ainwood]

Right. I'll demonstrate the method on a simpler example.

In the diagram below, basically you start at point "A" and you count the different ways to get to each of the next possible intersections. Obviously, you can either go down, or right.

Going down, there is only 1 possible way of getting to that intersection, so we write "1" there.

Going across, there is also only 1 possible way to get to the first intersection, so we write 1 there as well.

With the rule stating that you can only travel right or downwards, it is fairly obvious that there is only one possible way of getting to any of the intersections on the left-hand edge. We can write "1" by each of these. By the same logic, we can write "1" at each intersection across the top.


Now for the internal intersections. This is a simple addition problem: At each intersection, the number of ways to get to it is the sum of the number of ways to get to the intersection on the left , and the number of ways to get to the intersection above it.

There is therefore two possible ways to get to the center intersection, three ways to get to either the bottom-middle or Right-middle intersection. These can be summed to show six ways of getting to "B".

 

[Lucky]

Yes, easier indeed, but still a lot of work.
Mind to post a new question now that you told us how to solve that one?

 

[ainwood]

Originally posted by Lucky
Yes, easier indeed, but still a lot of work.

It took me longer to draw the diagram than to count the different ways! (I make it 378 by the way).

Anyway, next question:

Arrange the numbers 2, 4, 6, 8, 10, 12, 14, 16 and 18 into a three x three square, so that each horizontal row and vertical column add to the same number.

 

[AVN]

Ainwood,

One of the possible solutions is :

12 - 14 - 4
2 - 10 -18
16 - 6 - 8

This was easy.

 

[Lucky]

Here:

The overall sum is 90 so each row and column has to add to 30.

 2 + 12 + 16 = 30
18 +  4 +  8 = 30
10 + 14 +  6 = 30
 =    =    =   --
30   30   30 ||90

:D

 

[ainwood]

Originally posted by AVN
This was easy.

YEP!

Your go

(you pipped Lucky by a few hundred trillian periods of the radiation, which correlates to the transition between both Hyperfineniveaustructures (?) of the basic state of the atom Caesium 133)

 

[Lucky]

Actually I was faster! I posted at 6:46 my time, ainwood at 6:45. BUT he actually posted his answer at 6:53 my time, not earlier. My last edit was at 6:52.

I don´t mind though, go ahead, next question!

 

[AVN]

The next question is not too difficult, and it is civ3 related.

You are playing the Germans against 6 random opponents (out of 15).
How many combinations of opponents are there ?

PS : Lucky you were indeed quicker , but as you requested I take it on.

 

[Lucky]

Should be 5005 possibilities!?
If I didn´t sleep too tight in my Stochastics lessons.

 

[Hitro]

That should be right, Lucky.

15 choose 6 is 5005 I think. Or we did something completely wrong as it seemed to be quite easy...

 

[AVN]

You are both completely right.
In general people don't like stochastics, so therefore I gave a simple problem.
But I understand that you both know a lot about it, so be prepared for my following problem

But for now it's Lucky's turn.

 

[ainwood]

Originally posted by Lucky
Actually I was faster! I posted at 6:46 my time, ainwood at 6:45. BUT he actually posted his answer at 6:53 my time, not earlier. My last edit was at 6:52.

I don´t mind though, go ahead, next question!

OOPS! Sorry - I looked at the order they appeared on the thread. I didn't see the post before it was edited.

 

[Lucky]

Now, let´s see. I don´t have time to think of a really nice one so this will have to do!

Division of 2 complex numbers:

z1=-SQRT(3) + i
z2=-SQRT(8) + i*SQRT(8)

Calculate z3=z1/z2 ! What is the absolute value of z3 (|z|) and what is its phase (f)?
Write z3 in the cartesian and the exponential form (z=a+ib, z=|z|*e^if)!

i is the imaginary unit!
I hope the translation was correct.