MT2_1997S sln

MT2_1997S sln - All 5 questions on Midterm 2 were worth 20...

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Question 1 ---------- Among a group of 3 cats (Tabby, Leo, and Garfield) and 4 dogs (Snoopy, Odie, Fido, and Bowser) we wish to arrange a photograph of 5 of them. How many ways are there to arrange 5 of the animals in a photograph under the following conditions (each set of conditions is independent of the others): 1) No conditions; any 5 animals may appear in any order. 2) There must be exactly 3 dogs in the photograph. 3) A cat must be in either the leftmost position or the rightmost position. Hint: Use inclusion-exclusion. Answer to Question 1: --------------------- (1) Worth 7 points. There are 7 choices for the leftmost position, 6 remaining choices for the next position, 5 for the next, 4 for the next, and 3 for the rightmost position. By the product rule, we have P(7,5) = 7*6*5*4*3 = 2520 arrangements. Little partial credit was given for errors here. Three points were awarded for calculating C(7,5) instead of P(7,5). (2) Worth 6 points. There are C(4,3) = C(4,1) = 4 ways to choose 3 dogs for the picture. Then there are 5*4*3 ways to choose positions for them in the picture. Then we have to choose 2 cats for the other 2 spots in the picture. There are C(3,2) = 3 ways to do this, and then 2 ways to arrange the cats in the 2 vacant spots left in the picture. Multiplying all of these choices together, we get 4*(5*4*3)*3*2 = 1440. Partial credit was given only in very special cases, because there is little work to show other than writing down the factors in the multiplication. (3) Worth 7 points. We use inclusion-exclusion. Let S be the set of all arrangements with a cat in the leftmost position, and let T be the set of all arrangements with a cat in the rightmost position. We want to find |S U T|. By inclusion-exclusion, |S U T| = |S| + |T| - |S intersect T|. Now we find |S|. There are 3 choices for a cat in the leftmost position. The remaining spots of the photo can be filled out without restriction. So there are 6*5*4*3 ways do accomplish this. Hence, |S| = 3*6*5*4*3 = 1080. By symmetry, |T| = 1080. Finally, we must calculate |S intersect T|, the number of arrangements with cats in the left and right-most positions. There are 3 ways to choose a cat for the leftmost position. Then 2 choices remain for a cat in the rightmost position. The middle 3 spots can then be filled out without restriction in 5*4*3 ways. So we get a total of |S intersect T| = 3*2*5*4*3 = 360. Hence, |S U T| = 1080 + 1080 - 360 = 1800.
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This note was uploaded on 03/08/2010 for the course MATH 55 taught by Professor Strain during the Spring '08 term at Berkeley.

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MT2_1997S sln - All 5 questions on Midterm 2 were worth 20...

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