By Khee Meng Koh, Eng Guan Tay
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Extra info for Counting: Solutions Manual
1 has 7 choices to pair with. After #1 has been paired oﬀ, the lowest number among those remaining chooses someone to pair oﬀ with. He has 5 choices. At each step, the lowest number remaining chooses someone to pair oﬀ with. This method ensures that there is no double-counting. Thus, the number of ways = 7 × 5 × 3 × 1 = 105. 6 There are three boys and two girls. (i) Find the number of ways to arrange them in a row. (ii) Find the number of ways to arrange them in a row so that the two girls are next to each other.
If there exists a one-to-one mapping f : A → B, then |A| ≤ |B| . 1) The Bijection Principle (BP) Let A and B be ﬁnite sets. If there exists a bijection f : A → B, then |A| = |B|. 1 (a) Find the number of positive divisors of n if (i) n = 31752; (ii) n = 55125. (b) In general, given an integer n ≥ 2, how do you ﬁnd the number of positive divisors of n? Solution (a) (i) Observe that 31752 = 23 × 34 × 72 . Thus a positive number z is a divisor of 31752 if and only if it is of the form z = 2a × 3b × 7c , where a, b, c are integers such that 0 ≤ a ≤ 3, 0 ≤ b ≤ 4 and 0 ≤ c ≤ 2.
There are 8! ways of doing this. To ensure that no two of A, B and Z are adjacent, we now put each of them one by one into the spaces between the other 8 persons (see ﬁgure below). Number of choices for A = 9. Number of choices for B = 8. Number of choices for Z = 7. 9 · 8 · 7. (xi) First, “tie” the 6 boys together, with Z adjacent to A, as a single block. e. ZAXXXXX or XXXXXAZ. e. the block and the other 4 men, to arrange in a row. There are 5! ways of arranging 5 entities. “Untie” the 5 boys and arrange them.
Counting: Solutions Manual by Khee Meng Koh, Eng Guan Tay