#### davidleung808

##### New member

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- Thread starter davidleung808
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What have you learned about permutations and combinations? Which is this?

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How many ways you can

CHOOSE

4 books out of 80 books?

Have you studied permutation/combination?

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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.80!/(80!-4!)=number of permutations. Correct?

"Number of ways" is a little vague.

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Ok. So what was the typoNo. I think there's both a (serious!) typo, and a wrong choice of formula here. The bag probably doesn't care about order.

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But that's still the wrong formula.

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It would seem that I got trigger happy with the factorial symbol. Sorry.

But that's still the wrong formula.

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?"

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But we're still waiting to hear more from the OP, so the question may be moot.

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

\(\displaystyle no. \ of \ combinations \ = \ \frac{n!}{(r!) * (n-r)!}\)

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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.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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Isn't that backward?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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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??

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Yes.

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??