Counting Numbers

• Sep 6th 2011, 05:22 PM
engr
Counting Numbers
I'm trying to get a handle on "counting" numbers and outcomes of an event. I am not sure if my math is right, so perhaps someone can show me where I went wrong (unless it's right..).

A hat contains 20 consecutive numbers (1 to 20). If four numbers are drawn at random, how many ways are there for the largest one to be a 16 and the smallest number to be a 5?

The way I solved this was to break it up into four components. Based on the wording of this problem, I am assuming that replacement does not occur.

So, four positions:

____ ____ ____ ____

Position one is reserved for the "16", so there is one possible outcome. Position two is reserved for the "5", so there is one possible outcome there as well. Positions three and four are reserved for the numbers between 5 and 16, with the second position having one less available number since position 3 has "used" that number.

So, my math looks like the following:

n = (1)(1)(10)(9)
n = 90 possible outcomes

Does that sound correct? Is my logic correct?
• Sep 6th 2011, 05:30 PM
Plato
Re: Counting Numbers
Quote:

Originally Posted by engr
A hat contains 20 consecutive numbers (1 to 20). If four numbers are drawn at random, how many ways are there for the largest one to be a 16 and the smallest number to be a 5?

There are 10 integers greater than 5 and less than 16.
So $\displaystyle \binom{10}{2}$.
• Sep 6th 2011, 05:41 PM
engr
Re: Counting Numbers
So, evaluating $\displaystyle \binom{10}{2}$ yields a final solution of 45 possible outcomes, rather than my initial 90.

p = $\displaystyle \frac{n!}{k!(n-k)!}$

p = $\displaystyle \frac{10!}{2!(10-2)!}$

Is there an easy explanation as to where I went wrong? Maybe this is naive, but I'm trying to get a better "gut" grasp on how these things work. Thanks!
• Sep 6th 2011, 05:51 PM
Plato
Re: Counting Numbers
Quote:

Originally Posted by engr
So, evaluating $\displaystyle \binom{10}{2}$ yields a final solution of 45 possible outcomes, rather than my initial 90.
p = $\displaystyle \frac{n!}{k!(n-k)!}$
p = $\displaystyle \frac{10!}{2!(10-2)!}$
Is there an easy explanation as to where I went wrong? Maybe this is naive, but I'm trying to get a better "gut" grasp on how these things work. Thanks!

Well if you pick 7 and then 10 that is the same as picking 10 and then 7.
Order makes no difference here.