Midterm1_2007_review_v2

Midterm1_2007_review_v2 - 1-1COMP170 – Fall 2007Midterm 1...

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Unformatted text preview: 1-1COMP170 – Fall 2007Midterm 1 Review2-1Question 1nguests are arranged seats in a row facing the audiencea) How many different ways are there to seat the n guestsat n seats?2-2Question 1Note that the order is important since the guests could belisted from left to right from the audience’s perspectiven!nguests are arranged seats in a row facing the audiencea) How many different ways are there to seat the n guestsat n seats?2-3Question 1nguests are arranged seats in a row facing the audienceb) Letn >2. How many ways to seat n guests if twospecific guests will not sit next to each other?2-4Question 1nguests are arranged seats in a row facing the audienceb) Letn >2. How many ways to seat n guests if twospecific guests will not sit next to each other?# waysnot sitting together=n!-# ways sitting together2-5Question 1nguests are arranged seats in a row facing the audienceb) Letn >2. How many ways to seat n guests if twospecific guests will not sit next to each other?# waysnot sitting together=n!-# ways sitting togetherTreating the two people as one indivisible group, there are(n-1)!different ways of seating the guests.Withinthe two-person group, there are 2 ways to sit them.# ways sitting together =2(n-1)!# waysnotsitting together =n!-2(n-1)!=(n-2)(n-1)!2-6Question 1nguests are arranged seats in a row facing the audiencec)n >10. Guests include 5 couples. For each couple,husband must sit with wife. How many seatings are there?2-7Question 1nguests are arranged seats in a row facing the audiencec)n >10. Guests include 5 couples. For each couple,husband must sit with wife. How many seatings are there?Treat each couple as an indivisible group, and others asindividual groups.There are(n-5)!ways to seatn-5groups.For each couple (group),there are 2 ways to seat husband and wife.# ways to seat guests =25(n-5)!3-1Question 2Zn={, ..., n-1}a) How many 5-element subsets ofZ10contain at least oneelement inZ3?3-2Question 2# 5-element sets ofZ10=(105)# 5-element sets ofZ10notcontaining elements ofZ3=(75)# 5-element sets ofZ10containingat least oneelement ofZ3Zn={, ..., n-1}a) How many 5-element subsets ofZ10contain at least oneelement inZ3?=105-753-3Question 2b) How many 5-element subsets ofZ10contain two odd andthree even numbers?3-4Question 2b) How many 5-element subsets ofZ10contain two odd andthree even numbers?Z10={,1,2,3,4,5,6,7,8,9}(52)ways to choose 2 numbers from{1,3,5,7,9}(53)ways to choose 3 numbers from{,2,4,6,8}By the product principle, we have(52)·(53)3-5Question 2c) Let n be positive and odd. Show that # of even-sizedsubsets ofZnequals # of odd-sized subsets ofZn3-6Question 2c) Let n be positive and odd. Show that # of even-sizedsubsets ofZnequals # of odd-sized subsets ofZnLetOn={1,3, .5, ..., n}Since(nk)=(nn-k), we haveXk∈Onnk=Xk∈Onnn-k=Xk∈EnnkEn={,2,4, ..., n-1}f(k) =n-kdefines a bijection betweenOnandEn3-7Question 2c) Let n be positive and odd. Show that # of even-sizedsubsets ofZnequals # of odd-sized subsets ofZnAlternatively:2n= (1 + 1)n=X≤i≤nni!...
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Midterm1_2007_review_v2 - 1-1COMP170 – Fall 2007Midterm 1...

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