Text: Brualdi, Introductory Combinatorics 5th Ed. Prof: Paul Terwilliger Selected solutions II for Chapter 2

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Math 7 Text: Brualdi, Introductory Combinatorics th Ed Prof: Paul Terwilliger Selected solutions II for Chapter 0 We proceed in stages: The answer is! pick gender to the parent s right order the girls clockwise! order the boys clockwise! Now assume that there are two parents, labelled P and Q Suppose we move clockwise around the table from P to Q Let n denote the number of seats between the two Thus n = 0 resp n = 0 if Q sits next to P at P s left resp right We now partition the set of solutions according to the value of n n # seatings 0!!!!!!! 7! 8! 9! 0! The answer is the sum of the entries in the right-most column, which comes to 0! 8 We make a change of variables Define y = x, y = x, y = x +, y = x 8 Note that {x i } is a solution to the original problem if and only if y 0, y 0, y 0, y 0, y + y + y + y = Therefore the number of solutions is + = 8 9 a There are 0 ways to choose six sticks from the twenty available sticks

b Label the sticks,,, 0 Suppose we choose six sticks labelled {x i } with x < x < < x Define y = x, y = x x, y = x x, y = x x, y = x x, y = x x, y 7 = 0 x Observe that the solutions {x i } to the original problem correspond to the integral solutions {y i } 7 for y i 0 i 7, 7 y i = 9 Therefore the number of solutions {x i } to the original problem is 9 + 7 = 7 c We proceed as in b with the modification y = x, y = x x, y = x x, y = x x, y = x x, y = x x, y 7 = 0 x The solutions {x i } to the original problem correspond to the integral solutions {y i } 7 for y i 0 i 7, 7 y i = Therefore the number of solutions {x i } to the original problem is + 7 0 = 7 We proceed in stages: The answer is We proceed in stages: hand out the orange give one apple to each of the other children distribute remaining 0 apples to children hand out lemon drink hand out lime drink give one orange drink to each of the remaining two students distribute remaining 8 orange drinks to students

The answer is the product of the entries in the right-most column, which comes to a For i let x i denote the number of books on shelf i We seek the number of integral solutions to x i 0 i, x i = 0 The answer is 0 + = b We proceed in stages: put book on a shelf put book on a shelf 0 put book 0 on a shelf The answer is the product of the entries in the right-most column, which comes to 0 c We proceed in stages: order the books 0! pick a solution {x i } to part a put the first x books on shelf put the next x books on shelf put the next x books on shelf put the next x books on shelf 7 put the last x books on shelf The answer is the product of the entries in the right-most column, which comes to 0! a The word has 7 letters with repetitions letter A B D E H I K O P R S T mult

7!!! b The word has 9 letters with repetitions letter A C F H I L N O P T U mult 9 c The word has letters with repetitions 9!!!! 9! letter A C E I L M N O P R S T U V mult 9 d!!!!!! 9! 0 The number of ways to pick the bagels is equal to the number of integral solutions for x i 0 i, x i = which comes to + = 0 This is the denominator We now compute the numerator for the first probability The number of ways to pick the bagels so that you get at least one bagel of each kind is equal to the number of integral solutions for y i 0 i, y i = 9 which is 9+ = The first desired probability is 0

We now compute the numerator for the second probability The number of ways to pick the bagels so that you get at least three sesame bagels is equal to the number of integral solutions for z i 0 i, z i = which is + = 7 The second desired probability is 7 0 The sample space S satisfies The first event E satisfies S = 9! E = 9! 9 The first desired probability is E / S which comes to 9 The second event F satisfies F =!! The second desired probability is F / S which comes to!! 9! 9 The size of the sample space is This is the denominator for a e below a The die numbers must be some permutation of,,, ways or,,, ways The numerator is + = 0 The desired probability is 0/ b One dot occurs either once ways or twice ways or not at all ways The desired probability is + + c The desired probability is / d The desired probability is P, / e The die numbers must be some permutation of i, i, i, j ways or i, i, j, j ways Here we mean i j The desired probability is +