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it was for #54;; sorry | it was for #54;; sorry | ||
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| + | [[Category:MA375Spring2009Walther]] | ||
| + | If you were to use the equation in the book though, I found myself getting very confused with it and it not working. Thus I had to just manually write out all of the combinations. Can someone explain how to use the given formula? | ||
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| + | Thanks, | ||
| + | Kristen | ||
Latest revision as of 18:13, 12 February 2009
Back to HW 4: 5.5
I tried using the equation listed in the book, but I ended up with 0 in the denominator, am I missing something?
Using the equation on page 378 I got 41.--Spfeifer 20:37, 11 February 2009 (UTC)
For this problem, I started looking at it as five indistinguishable objects into three indistinguishable boxes and then added the distinguishable part back in for the objects. You could have (5,0,0) (4,1,0) (3,2,0) (3,1,1) (2,2,1). Adding the distinguishable factor back in, for (5,0,0) there is 1 option, for (4,1,0) there is (5 choose 4)x(1 choose 1), for (3,2,0) there is (5 choose 3)x(2 choose 2), for (3,1,1) there is (5 choose 3)x(2 choose 1)x(1 choose 1), and for (2,2,1) there is (5 choose 2)x(3 choose 2)x(1 choose 1). Add these together and my answer was 66.
I'll verify that the response about the equation on page 378 is correct. As for getting a zero in the denominator, remember that 0! = 1. -NF
(5,0,0) (4,1,0) (3,2,0) (3,1,1) (2,2,1). Does the order matter for this problem?? I wasn't sure what to do
it seems to me that there are only five ways to arrange it..
it was for #54;; sorry
If you were to use the equation in the book though, I found myself getting very confused with it and it not working. Thus I had to just manually write out all of the combinations. Can someone explain how to use the given formula?
Thanks, Kristen
