Midterm1_2008_review_v2

Midterm1_2008_review_v2 - 1-1COMP170 – Fall 2008Midterm 1...

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Unformatted text preview: 1-1COMP170 – Fall 2008Midterm 1 Review2-1Question 18 men and 8 women are invited to a party at which theyare seated at a long rectangular tablea) How many different ways are there to seat the n guests at n seats?2-2Question 1There are8!ways to seat the men and8!ways of seating the women.By the product principle,(8!)2= 1,625,702,4008 men and 8 women are invited to a party at which theyare seated at a long rectangular tablea) How many different ways are there to seat the n guests at n seats?2-3Question 18 men and 8 women are invited to a party at which theyare seated at a long rectangular tableb) Suppose one man and one woman will not sit across from eachother. How many ways are there to seat all of the men and women?2-4Question 18 men and 8 women are invited to a party at which theyare seated at a long rectangular tableb) Suppose one man and one woman will not sit across from eachother. How many ways are there to seat all of the men and women?SupposeW1andM1quarelled.There are8!ways of seating the men; then7possible locations toseatW1, and after that,7!ways of seating the remaining 7 women.By the product principle,8!×7×7! = 1,422,489,6002-5Question 18 men and 8 women are invited to a party at which theyare seated at a long rectangular tablec) Suppose if a man and a woman are married, they sit exactly acrossfrom each other. If there are 3 married couple, how many ways arethere to seat everyone?2-6Question 18 men and 8 women are invited to a party at which theyare seated at a long rectangular tablec) Suppose if a man and a woman are married, they sit exactly acrossfrom each other. If there are 3 married couple, how many ways arethere to seat everyone?There are8!ways of seating the men. After the men are seated, the3 wives must sit across from their husbands. The remaining 5 womenhave5!seating arrangements. So, by the product principle,8!×5! = 4,838,4003-1Question 2Give a combinatorial proof of the identity, for alln≥11.n3n-35n-83=n3n-33n-65Note: An algebraic proof of this identity will not be accepted as asolution.3-2Question 2Consider the problem of how to colornitems so that3arered,5aregreen,3areblueand the remainingn-11are yellow.The left hand side obviously counts thisConsider the problem of how to colornitems so that3arered,3areblue,5are green and the remainingn-11are yellow.The right hand side oObviously counts thisNotice that both problems are the same.We only change the order in which we do the colorings.Give a combinatorial proof of the identity, for alln≥11.n3n-35n-83=n3n-33n-65Note: An algebraic proof of this identity will not be accepted as asolution.4-1Question 3Starters (A List)10 Types,A1,A2,...,A10Main Courses (B List)15 Types,B1,B2,...,B15(a) How many different menus (3from A,2from B) can you create?...
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This note was uploaded on 08/25/2010 for the course COMP COMP170 taught by Professor M.j.golin during the Spring '10 term at HKUST.

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Midterm1_2008_review_v2 - 1-1COMP170 – Fall 2008Midterm 1...

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