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**gracenalec****Member**- Registered: 2006-01-23
- Posts: 4

Hi guys,

I have a question here which is really causing me headache. Hopefully, someone could help me with this

The positions of nine trees which are to be planted along the sides of a road, five on the north side and four on de south side.

O O O O O ---North

O O O O ---South

So, the first question is:

Find the number of ways in which this can be done if the trees are all of different species.

ANS: what i have in mind is this... 9P5 x 4! coz first one gotta pick 5 out of the nine to arrange 1st and then the balance will be 4!

next question will be:

If the trees in the above question are planted in random, find the probability that two particular trees are next to each other on the same side of the road.

ANS: the prob I have is that I dunno if i should treat each tree as unique or not. So far, what I have in mind is this:

(4! x 2! x 4!)/(9P5 x 4!) + (3! x 2! x 5!)/(9P5 x 4!)

I'm nt sure if this is right or not...

Pls do correct me if i'm wrong..THANKS A MILLION!!!

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

In the first Q it says "all of different species" if that mean 9 different species they may as well be in a row, and the answer will be: **9!**

Which = 9P5 × 4! anyway (9P5 × 4! = (9!/(9-5)!) × 4! = (9!/4!) × 4! = 9!)

Hopefully someone else can solve the other Q for you.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**kempos****Member**- Registered: 2006-01-07
- Posts: 77

i would do it this way

you pick out the first tree - there are 9 posibilities and plant it on north or south - 2

so the first tree 9x2 ways

second 8x2

third; 7x2

fourth 6x2

fifth 5x2

sixth 4x2

seventh 3x2

eight 2x2

ninth 1x2

9x2x8x2x7x2x6x2x5x2x4x2x3x2x2x2x2 so 9!x2^10

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**kempos****Member**- Registered: 2006-01-07
- Posts: 77

question 2 (but here i'm not sure, but will try):

probability that the first is on the north: 1/2

probability that the second is on the north is 1/2

probability that they are both on the north 1/4

on the north we have 5 trees: 5!

they have to be next to each other (it's like one tree): 4!

the can also change sides (one is on the left of another and vice versa) 2!

probability that they are on the northand next to each other: 1/4*(4!*2!)/5!=1/20

on the south: 1/4*(3!*2!)/4! = 1/8

north or south and next to each other: 1/20+1/8 = 2/40+5/40=7/40

but as i said i'm not sure

*Last edited by kempos (2006-01-24 03:00:37)*

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Maybe I'm missing something here, but I think this way works.

Either a tree is on the end of a row, in which case there's a 1/8 chance of a specific tree being next to it, or a tree is in the middle of a row somewhere, in which case there's a 2/8 = 1/4 chance of a specific tree being next to it.

So, the total probability is 4/9*1/8 + 5/9*2/8 = 4/72 + 10/72 = 14/72 = 7/36.

Why did the vector cross the road?

It wanted to be normal.

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**jkiller****Member**- Registered: 2006-01-24
- Posts: 2

i am not sure about my answer

the 2 particular tree are next to each other = 2way (north or south)

north 2! * 4! + 2! * 3!

means 2*2*24*2*6 = 1152

i know is a bit simply, just my comments about it...sorry if i cant help

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**kempos****Member**- Registered: 2006-01-07
- Posts: 77

guys in my opinion we also have to remember that the trees have to be put on the same side of the road. here probablity is 1/2 * 1/2 = 1/4

in my opinion this is what we should start with.

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**kempos****Member**- Registered: 2006-01-07
- Posts: 77

jkiller you were finding the number of ways not the probability

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**kempos****Member**- Registered: 2006-01-07
- Posts: 77

mathsyperson, you are saying that there are nine trees in the row which is not the truth!! remember that we have to choose the correct side first

*Last edited by kempos (2006-01-24 03:08:28)*

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

kempos wrote:

mathsyperson, you are saying that there are nine trees in the row which is not the truth!! remember that we have to choose the correct side first

No I'm not. There are four trees that are on the edge of their row, and five trees that are somewhere in the middle of their row.

Why did the vector cross the road?

It wanted to be normal.

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**jkiller****Member**- Registered: 2006-01-24
- Posts: 2

yaya....i did in how many way not probability...hehe..sorry...let me think again....

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**gracenalec****Member**- Registered: 2006-01-23
- Posts: 4

hurm...mathsyperson..i dun really get ur way...but i was thinking

(2! x 7!)/9! according to the probability formular = n(A)/n(S)

what u guys think?

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**gracenalec****Member**- Registered: 2006-01-23
- Posts: 4

umm..guys, what if there are 2mangnolias, 3 figs and 4 prunes trees. What is the probability that 2 magnolias are on the opposite sides of the road. (1 must be north, another must be south)...any idea??

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

If the first tree is on the top row and the second is on the bottom, then the total probability is 5/9*4/9 = 20/81.

However, the position of the trees could also be reversed, so this value must be doubled.

**Probability: 40/81**

Why did the vector cross the road?

It wanted to be normal.

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**kempos****Member**- Registered: 2006-01-07
- Posts: 77

probability that the firsy magnolia is north is 1/2

probabbility that the second is south 1/2

different side: 1/4

reversed *2

final: 1/2

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**pranshu****Member**- Registered: 2006-02-21
- Posts: 3

first que ans will be 9! as first will have 9 options ,second will have 8 and so on.

second que ans will be 8! x 2/ 9! as two trees can be planted together in 8! x 2 ways. it can be done by assuming two trees as one then we can do this in 8! ways and their places can be interchanged hence should be multiplied by 2.

hope my solution is satisfactory.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I just made a discover that has to do with this problem.

If you connect a counting number of object in a line and figure out how many

are beside one another, it is also true that if you add one more object to the line and

link it to the other end forming a circle, now the pairing probability is the SAME!!!!!

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Boy, that was not explained very well.

You know the probability of a certain chosen two distinct objects being beside one another.

If it in a row, that probability is the same as a circular linked list if you add in the missing link.

**igloo** **myrtilles** **fourmis**

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**pranshu****Member**- Registered: 2006-02-21
- Posts: 3

budy if one wants to arrange n objects in a circular positions symmetrickely then he can do it in n-1! ways which is this case as we can diffrenciate between clockwise and anti clockwise because of diffrent species of trees.

with this logic the

ans 1 will be 8!

and ans 2 will be 7!x 2/ 8!

check it up n then write u r comment and if u r ans is not this then if possible give your answer in detail so that i can correct my approach.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Okay, 14/81 is the answer to # 2.

The question being: If the trees in the above question are planted in random, find the probability that two particular trees are next to each other on the same side of the road.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Also if 17 people are standing in a row and you are one of them.

Then the probability that you are beside a certain person is 2/17.

If 18 people are in a circle and you are one of them, then the

probability you are beside a certain person is still 2/17.

Pretty cool, huh? I just discovered this yesterday. The missing link.

**igloo** **myrtilles** **fourmis**

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

So, if there are three people in a circle, the probability you are beside a certain person is 2/2. Yeah!

This could be adapted to one of those "I bet you" things, where the gambler knows the odds better than other people.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

So anyway, how I reason 14/81 for the 4 trees and 5 trees across the street, and the chance two are beside each other is:

probability of 1. plus probability of 2., where

1. On four trees side, beside each other is: (2/4)(4/9)(3/9), but perhaps

I am wrong. and

2. On five trees side, beside each other is: (2/5)(5/9)(4/9)., but this may

be wrong, now that I think about it.

The 2/4 and the 2/5 multiplier is correct because if the two trees are on the

same side, then that is the probability they are beside each other.

The other two factors (4/9)(3/9) was my guess for the chance that the

two trees are on the 4 tree side of the street, but I bet I'm wrong with

that part.

Sorry for any confusion saying I had the right answer.

**igloo** **myrtilles** **fourmis**

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