Confusing probability problem.

A gardener plants 7 maple trees and 5 birch trees in a row. You may assume that all trees are distinguishable from one another. She plants them in a random order, each arrangement being equally likely. Let m/n, in lowest terms, be the probability that no two birch trees are next to one another. Find m+n.

Re: Confusing probability problem.

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Originally Posted by

**wannabebrainiac** A gardener plants 7 maple trees and 5 birch trees in a row. You may assume that all trees are distinguishable from one another. She plants them in a random order, each arrangement being equally likely. Let m/n, in lowest terms, be the probability that no two birch trees are next to one another. Find m+n.

The probability that no two birch trees are next to one another is:

$\displaystyle \dfrac{\dbinom{8}{5}(5!)(7!)}{12!}$

The numerator is the number to plant so that no two birch trees are next to one another.

Re: Confusing probability problem.

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Originally Posted by

**Plato** The probability that no two birch trees are next to one another is:

$\displaystyle \dfrac{\dbinom{8}{5}(5!)(7!)}{12!}$

The numerator is the number to plant so that no two birch trees are next to one another.

Alright, thanks! However, what did you do exactly?

Re: Confusing probability problem.

Quote:

Originally Posted by

**wannabebrainiac** Alright, thanks! However, what did you do exactly?

__M__M__M__M__M__M__M__

The maples create eight places to plant the five birch trees that no two are next to one another.

Choose five of them. Arrange the birch, 5!, arrange the maples, 7!

Divide by the total.

Re: Confusing probability problem.

I would have explained it a little bit differently:

Quote:

Originally Posted by

**Plato** __M__M__M__M__M__M__M__

The maples create eight places to plant the five birch trees that no two are next to one another.

I like this approach - it gives C(8,5) for the numerator. That's the number of ways you can arrange the trees so no two birches are next to each other.

Quote:

Originally Posted by

**Plato** Choose five of them. Arrange the birch, 5!, arrange the maples, 7!

Divide by the total.

My explanation is: for the denominator you need the total number of ways the trees can be arranged. Consider the birches - you have 5 of them and they can go into any of 12 spots, so the number of possible combinations is C(12,5). Or you could consider the 7 maples - they could be positioned in any of C12,7) ways, which is the same thing. So the probability of getting no two birches together is C(8,5)/C(12,5).

I'm not saying my explanation is necessarily better, but for me at least I think it's a bit more intuitive.

Re: Confusing probability problem.

Hello, wannabebrainiac!

Quote:

A gardener plants 7 maple trees and 5 birch trees in a row.

You may assume that all trees are distinguishable from one another.

She plants them in a random order, each arrangement being equally likely.

Let $\displaystyle \frac{m}{n}$, in lowest terms, be the probability that no two birch trees are next to one another.

Find $\displaystyle m+n.$

There are $\displaystyle 12!$ possible arrangements of the trees.

Place the maples in a row, inserting a space before, after and between them.

. . $\displaystyle \_\:m\:\_\:m\:\_\:m\:\_\:m\:\_\:m\:\_\:m\:\_\:m\: \_ $

There are $\displaystyle 7!$ arrangments of the seven maple trees.

Select five of the eight spaces and insert the birch trees.

. . There are: .$\displaystyle _8P_5 \:=\:8\cdot7\cdot6 \:=\:336$ ways.

Hence, there are: .$\displaystyle 7!\cdot336$ ways in which the birches are not adjacent.

Therefore: .$\displaystyle P(\text{nonadjacent birches}) \:=\:\frac{7!\cdot336}{12!} \:=\:\frac{7}{99}$

. . and $\displaystyle m + n \:=\:106.$

This is the answer that Plato got.

I just approached the problem slightly differently.