# Probability of selecting 5 numbers from 1-40, no two being consecutive

• Jun 4th 2010, 06:42 AM
fourierwarrior
Probability of selecting 5 numbers from 1-40, no two being consecutive
Supposing we select 5 integers at random from 1 to 40 inclusive.

How could we find the probability of none of these numbers being consecutive?

I know we have 40 choose 5 outcomes total. I'm having trouble figuring out how many ways two or more of the 5 numbers would be consecutive though.
• Jun 4th 2010, 06:47 AM
Plato
Quote:

Originally Posted by fourierwarrior
Supposing we select 5 integers at random from 1 to 40 inclusive. How could we find the probability of none of these numbers being consecutive?
I know we have 40 choose 5 outcomes total. I'm having trouble figuring out how many ways two or more of the 5 numbers would be consecutive though.

How many ways to arrange 35 zeros and 5 ones so that no two ones are adjancent?
• Jun 4th 2010, 07:40 AM
fourierwarrior
Ok, so there's 40! ways of ordering the 5 1's and 35 0's total.

I guess we could group the consecutive 1's as follows:

5 1's [11111][000....00]

35 ways

4 1's [01111][010....00]

5*35 ways??

3 1's [11100][010...10]

This way seems really difficult, I'm not really sure where to start.
• Jun 4th 2010, 07:56 AM
Plato
Here is a hint.
How many ways to arrange 11100000 with no adjacent ones?
_0_0_0_0_0_
There are five zeros which create six places to put the ones.
So the answer is $_6C_3=\frac{6!}{3!\cdot 3!}$.
You appply that to this problem.
• Jun 4th 2010, 08:10 AM
fourierwarrior
Ohhhh! That makes sense.

SO there are 35 zeros, which leaves 36 places to put the ones, so there are 36C5 ways to order the ones among the zeros.

Hence, the probability I'm looking for is 36C5 / 40C5.

Thank you!
• Jun 4th 2010, 08:13 AM
Plato
Quote:

Originally Posted by fourierwarrior
SO there are 35 zeros, which leaves 36 places to put the ones, so there are 36C5 ways to order the ones among the zeros.
Hence, the probability I'm looking for is 36C5 / 40C5.

You have it.