# Counting: Solutions Manual by Khee Meng Koh, Eng Guan Tay

By Khee Meng Koh, Eng Guan Tay

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Additional resources for Counting: Solutions Manual

Example text

For example, if r = 3, f ({1, n − 1, n}) = {2, 3, . . 6 The number 4 can be expressed as a sum of one or more positive integers, taking order into account, in the following 8 ways: 4=4 =1+3 =3+1 =2+2 January 16, 2013 14:10 9in x 6in Counting: Solutions Manual (2nd Edition) The Bijection Principle b1502-ch05 45 =1+1+2 =1+2+1 =2+1+1 = 1 + 1 + 1 + 1. Show that every natural number n can be so expressed in 2n−1 ways. Solution We write 4 = 1 + 1 + 1 + 1 and note that there are three spaces between two “1”s ﬁlled up by “+”s in the expression.

4! ) + · · · + ((n + 1)! ) = (n + 1)! − 1. January 16, 2013 14:9 9in x 6in Counting: Solutions Manual (2nd Edition) This page intentionally left blank b1502-ch03 January 16, 2013 14:10 9in x 6in Counting: Solutions Manual (2nd Edition) b1502-ch04 4 Applications Principle of Complementation (CP) Let A be a subset of a ﬁnite set B. Then |B\A| = |B| − |A|, where B\A = {x | x ∈ B but x ∈ / A}, called the complement of A in B. 1 There are 6 boys and 5 men waiting for their turn in a barber shop.

We then consider the remaining 5 consonants. Let us imagine tentatively that these 5 consonants are identical, and they are to be placed in 5 distinct boxes as shown in the ﬁgure below so that boxes (2), (3) and (4) are 51 January 16, 2013 14:10 9in x 6in Counting: Solutions Manual (2nd Edition) b1502-ch06 Counting: Solutions Manual 52 not vacant (since no two of I, A, O and E are adjacent). To meet this requirement, we place one each in boxes (2), (3) and (4). Then the 2+5−1 6 = 4 remaining two can be placed freely in the boxes in 5−1 ways.