# Thread: the last 2 probability questions

1. ## the last 2 probability questions

give me hints so i will solve them by myself ...
Thank you a lot

2. Originally Posted by Narek
give me hints so i will solve them by myself ...
Thank you a lot
As Plato said before, please type them out - it's easier to quote and to follow the thread if there are more responses.

Nr. 46:
First: how many table arrangements with 8 people are there if the table were a long table (or if you want, a police line-up)?

Then, how many are the same if you take rotations at a round table into account?

Nr. 20:
In a polygon with n sides (n >= 3), a diagonal from edge-1 can go to which of the other edges?

3. ## hmmm ...

still i dont get it! maybe a little more help please?????

4. Originally Posted by Narek
still i dont get it! maybe a little more help please?????
Nr. 46:
When you have a line of 8 people, in how many ways can you arrange those 8 people? That's step 1.

Nr. 20:
Take for instance a hexagon, i.e., a polygon with n=6 sides. It also has 6 corners. Take one of those corners. To how many other corners can you draw lines? How many of those lines are called diagonals?

I draw it and I see that from any edge we can draw n/2 lines and from those only 1 is diagonal. I dont know how to show all these in mathematics language.

my problem about this question, is that I do NOT understand the questions even. its sad, but true. I dont understand what it says ...
is this solution true for this:

permulation WITHOUT repetition:

P(8,8) = 8! / (8-8)! = 40320
because of rotation we might have 40320 / 8 = 5040 different arrangements

True????

6. Originally Posted by Narek

I draw it and I see that from any edge we can draw n/2 lines and from those only 1 is diagonal. I dont know how to show all these in mathematics language.
No, a diagonal is any line from one corner to another corner which is not an edge.

Originally Posted by Narek

my problem about this question, is that I do NOT understand the questions even. its sad, but true. I dont understand what it says ...
is this solution true for this:

permulation WITHOUT repetition:

P(8,8) = 8! / (8-8)! = 40320
because of rotation we might have 40320 / 8 = 5040 different arrangements

True????
That's true, for the part of putting 8 people in a line.
Now if you put them in a circle, there is no "first" person in the line. So, by rotating, how many of permutations are the same arrangement?

7. Nr. 20)

so by mean of that, from every corner to others, there are n/2 lines. but there are lines which are repeated. how can i write this in mathematical language, and what to do with repeating lines?

Nr. 46)

P(8,8) = 8! / (8-8)! = 40320
because of rotation we might have 40320 / 8 = 5040 different arrangements

actually i divided the result by 8 so the problem goes around a circular table. I dont know what else should i do ... ?????

8. Originally Posted by Narek
Nr. 20)

so by mean of that, from every corner to others, there are n/2 lines. but there are lines which are repeated. how can i write this in mathematical language, and what to do with repeating lines?
Where did you get the formula n/2? From corner-2 there's an edge running to corner-1 and an edge to corner-3. The lines to the other corners are diagonals. That gives how many?

And how many times do you then count each line? Hint: a line has two endpoints...

Originally Posted by Narek
Nr. 46)

P(8,8) = 8! / (8-8)! = 40320
because of rotation we might have 40320 / 8 = 5040 different arrangements

actually i divided the result by 8 so the problem goes around a circular table. I dont know what else should i do ... ?????