Probability

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davidleung808

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Oprah has a recommended reading list of 80 books, but you can only fit four of them in your carry-on bag. How many ways can you select 4 of the 80 recommended books for your upcoming flight? 20.
 
Is 20 your answer? It seems a little small.

What have you learned about permutations and combinations? Which is this?
 
davidleung808 said:
Oprah has a recommended reading list of 80 books, but you can only fit four of them in your carry-on bag. How many ways can you select 4 of the 80 recommended books for your upcoming flight? 20.
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How many ways you can

CHOOSE

4 books out of 80 books?

Have you studied permutation/combination?
 
80!/(80!-4!)=number of permutations. Correct?
"Number of ways" is a little vague.
 
firemath said:
80!/(80!-4!)=number of permutations. Correct?
"Number of ways" is a little vague.
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No. I think there's both a (serious!) typo, and a wrong choice of formula here. The bag probably doesn't care about order.
 
I'm hoping you meant to say 80!/(80-4)! . Factorial doesn't distribute ... and what you typed evaluates to something very, very close to 1.

But that's still the wrong formula.
 
Dr.Peterson said:
I'm hoping you meant to say 80!/(80-4)! . Factorial doesn't distribute ... and what you typed evaluates to something very, very close to 1.

But that's still the wrong formula.
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It would seem that I got trigger happy with the factorial symbol. Sorry.
I meant to write what you said, (80-4) instead of (80!-40!). Bleh. But I don't know if OP wants permutations or combinations....."number of ways?"
 
There's some ambiguity about it, as to whether we are interested in the result (sets of books) or the process (choosing one at a time) of "ways to select"; but usually such a question means "how many possible subsets are there", since, as I said, a bag doesn't care about order. If it were described as having distinct compartments, it would be different.

But we're still waiting to hear more from the OP, so the question may be moot.
 
Where n = total number of elements and r = no. of elements to be picked from set of n

\(\displaystyle no. \ of \ permutation \ = \ \frac{n!}{(n-r)!}\)................. and

\(\displaystyle no. \ of \ combinations \ = \ \frac{n!}{(r!) * (n-r)!}\)
 
You can choose any of the 80 books first then any of the remaining 79 next, 78 third and, finally, 77 last. There are (80)(79)(78)(77)= 37957920 choices. That is the same as Subhotosh Khan's \(\displaystyle \frac{n!}{(n-r)!}\) for "combinations". Since the bag doesn't care what order the books are put in, we need to divide by 4!= 24 to remove the 24 different orders in which the four books can be put in. That gives \(\displaystyle \frac{37957920}{24}= 1581580\) choices. Again that is the same as Subhotosh Khan's formula for "permutations".[/tex]
 
HallsofIvy said:
Since the bag doesn't care what order the books are put in, we need to divide by 4!= 24 to remove the 24 different orders in which the four books can be put in. That gives \(\displaystyle \frac{37957920}{24}= 1581580\) choices. Again that is the same as Subhotosh Khan's formula for "permutations".[/tex]
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I would say: 37957920 choices includes 24 different orders for each 4 book selection. Therefore, we need to remove 23 out of every 24 choices by dividing the result by 24.
 
HallsofIvy said:
You can choose any of the 80 books first then any of the remaining 79 next, 78 third and, finally, 77 last. There are (80)(79)(78)(77)= 37957920 choices. That is the same as Subhotosh Khan's \(\displaystyle \frac{n!}{(n-r)!}\) for "combinations". Since the bag doesn't care what order the books are put in, we need to divide by 4!= 24 to remove the 24 different orders in which the four books can be put in. That gives \(\displaystyle \frac{37957920}{24}= 1581580\) choices. Again that is the same as Subhotosh Khan's formula for "permutations".
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Isn't that backward?
 
Yes, they are backwards. Thanks for the correction. The number of "combinations". which does not count permutations or the same letters as "different" is necessarily smaller than the number of "permutations" so has the division by r!.
 
Hi Dr. Petersen,

The order in which the four books are selected does not matter. This is a problem of selecting r =4 books from a group of n = 80. I am looking for the number of combinations of 80 books taken 4 at a time. I use the formula n! / {(n – r)! r!}= 80! / {(80 – 4)! 4!} = 80! / 76! 4! = 37957920 / (4*3*2*1) = 37957920 / 24 = 1581580

Thus, there are 1581580 ways to choose four books for a carry-on bag. Am I correct??
 
davidleung808 said:
Hi Dr. Petersen,

The order in which the four books are selected does not matter. This is a problem of selecting r =4 books from a group of n = 80. I am looking for the number of combinations of 80 books taken 4 at a time. I use the formula n! / {(n – r)! r!}= 80! / {(80 – 4)! 4!} = 80! / 76! 4! = 37957920 / (4*3*2*1) = 37957920 / 24 = 1581580

Thus, there are 1581580 ways to choose four books for a carry-on bag. Am I correct??
Click to expand...
Yes.
 
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