Discrete Mathematics with Graph Theory (3rd Edition) 218

Discrete Mathematics with Graph Theory (3rd Edition) 218 -...

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216 Solutions to Exercises 25. Both students are right and we can see this with several applications of formula (5), p. 350: (~) = (~=D + (nkl). Since (~~) + (~~) = (~~), X's answer is (~~) + (~~). Since (~~) = (~~) + (~~) and (~~) = (~~) + (~~), Y's answer is (~~) + (~~) - (~~) = (~~) + G~), in agreement with x. 26. The required probability is the sum (*) P(one head) + P(3 heads) + . .. + P(k heads) = (~) + (;) 2~··· + (~) h k { n if n is odd were = n -1 ifn is even. If n is odd (so k = n - 1), we can use the identity (;) = (;=D + (n~l) to rewrite the expression on the right of (*) as (n~l) + (n~l) + (n;-l) + (n3"l) + . .. + (~=~) + (~=D 2 n 2 n - 1 Problem 34 of Section 7.7 tells us this is ~ = ~. If n is even (so k = n), the identity (;) (;=D + (n~l) allows us to rewrite the right side of (*) as (n~l) + (n~l) + (n;-l) + (n3"l) + . .. + (~=~) + (~=~) + (~) 2 n Since (~) = (~=D, this is again 2;:1 = ~, giving the result. Exercises 8.1 1. [BB] To find the mid-point of a line segment
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Unformatted text preview: AB, choose a radius r (for example, r = lAB!) such that the arcs with centers A and B and radius r meet at two distinct points, P and Q. The point of intersection, M, of AB and PQ is the required mid-point. To see why this works, consider the diagram to the right, in which the labels are given as before and line segments AP, BP, AQ, BQ are joined. Note that LP AB = LP BA since 6.AP B is isosceles. Since D.BPQ is congruent to 6.APQ (three pairs of sides of equal length), LBPQ = LAPQ. Thus, triangles PAM and P BM are congruent (two equal pairs of sides and equal contained angles), so IAMI = IBMI as required. 2. Let A be any point of . Draw an arc with center A and radius r = IAPI. Let B be the intersection of this arc with t. Draw arcs with radius r and centers P and B meeting in A and C. Then the line through PC is the desired line. To see this, note that PABC has four equal sides, so it's a parallelogram. A ~-+--t=+----7 B...
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