MT2_1997S sln

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

This preview shows pages 1–2. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 03/08/2010 for the course MATH 55 taught by Professor Strain during the Spring '08 term at Berkeley.

### Page1 / 5

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online