Need Help with very difficult probability question! (choosing 5 from 11)

[johnnnyjoemc]

Question:
In how many ways can a committee of five be chosen from 11 people if two particular people will not work on the same committee (that is, if one is included, the other must be excluded)

Another question:
A girl chooses six numbers out of 45. 19 will be her third lowest number. How many ways are there for her to do this?

 

[Dr.Peterson]

Question:
In how many ways can a committee of five be chosen from 11 people if two particular people will not work on the same committee (that is, if one is included, the other must be excluded)

Another question:
A girl chooses six numbers out of 45. 19 will be her third lowest number. How many ways are there for her to do this?

These aren't "very" difficult; but they do require thought. What thought have you put into them? What have you learned that might help?

If you haven't already read our submission guidelines, please do (and then follow them), so that we have the information we need in order to help you.

For your first problem, assuming you need a hint to get started, you might consider counting separately all committees that include only one of the "particular people", and those that include neither. Or, you might count all possible committees, and subtract those that include both.

For the second, you might count ways to choose the two numbers lower than 19, and ways to choose the three numbers higher than 19.

These are typical ways to organize a problem like these.

 

[pka]

In how many ways can a committee of five be chosen from 11 people if two particular people will not work on the same committee (that is, if one is included, the other must be excluded)

Another question: A girl chooses six numbers out of 45. 19 will be her third lowest number. How many ways are there for her to do this?

Here is a second to think about the first one. If we know the number of ways to choose a committee that does include both of those people then we can easily find number of ways to choose a committee that does not include both of those people. How does that work?

 

[johnnnyjoemc]

My brain is wrecked over these questions... I still don't understand the method.

 

[Steven G]

My brain is wrecked over these questions... I still don't understand the method.

The 3rd number is 19.
So we have x x 19 y y y. Where x's are less than 19 and y's are bigger than 19. There are 45 numbers. We must make some assumptions on these numbers. If they are simply real numbers then the answer is infinity as each x has an infinite number of choices and the same for y's. So we will assume that the numbers are integers. Well do we start at 0 or 1 or does it even matter? Let's assume integers from 1 to 45. Let's also assume that numbers can't be repeated. (Note that this problem, in my opinion, is worded terribly!). Now there are 18 integers less than 19 of which we must pick 2 AND there are 26 integers above 19 of which we must pick 3.

You should be able to continue from here.

 

[Steven G]

My brain is wrecked over these questions... I still don't understand the method.

Lets look at the method pka mentioned.
Suppose there are 1000 ways to pick the committee if no restrictions at all--that is we are including the committees that include none of these two people, exactly one of the two and even both of them. Now let's say that there are 200 committees that include both of them. Well these 200 must be removed. So the result would be 1000 - 200 or 800. Unfortunately the 1000 and 200 are wrong. So let's get the correct values.

1st we want the number of ways to choose 5 from 11 (this would include the committees that have both people). This calculation should be routine.

Now we want to calculate the number of committees that both of them. Since we know that we are choosing these two for sure to be on the committee then we simply need to choose 3 (5-2=3) from the remaining 9 (11-2=9) people. This calculation is also routine.

Now subtract the two results.

 

[pka]

Question:
In how many ways can a committee of five be chosen from 11 people if two particular people will not work on the same committee (that is, if one is included, the other must be excluded)?

There is a total of \(\displaystyle \dbinom{11}{5}=462\) ways to choose 5 from 11.

Think this way. Suppose we go ahead and put the two adverse people on the committee and select three others.
That can be done in \(\displaystyle \dbinom{8}{3}=56\) ways. See here.

There are 462 ways to select a committee of five with no restrictions.
Of those there are 56 that contain both warring people.
So how many ways can we select a committee not containing both?

 

[dr.trovacek]

?

Should't the second claim be \(\displaystyle \binom{9}{3}\), since if we put two of the warring people on the committee then we have a group of 9 (11-2 = 9) people left to choose from?

 

[pka]

Should't the second claim be \(\displaystyle \binom{9}{3}\), since if we put two of the warring people on the committee then we have a group of 9 (11-2 = 9) people left to choose from?

Yes that is correct! Thanks.