Combinatorics: Problem Set 1 Math 50, Fall 2018 Jiayi Lou 1) Exercise 1, #2 (i) (ii) (iii) (i) Solution. The number of ways 12 students can be arranged in a row without restrictions 12 = 12!. is P12 (ii) Solution. We perform the following steps to arrange: · Step 1: Treating the 5 girls as one block and each boy as one block, arrange the blocks in a row. There are P88 possibilities. · Step 2: Arrange the 5 girls in their block. The number of possibilities is P55 . By the Multiplication Principle, the number of such arrangements is P88 P55 = 8!5!. (iii) Solution. We have the following steps: · Step 1: Arrange the 7 boys. There are P77 different ways. · Step 2: Put a girl into one of the 8 spaces in B1 B2 B3 B4 B5 B6 B7 , with 8 possibilities. · Step 3: Put a girl into one of the remaining 7 spaces, with 7 possibilities. .. . · Step 6: Put a girl into one of the remaining 4 spaces, with 4 possibilities. By the Multiplication Principle, the number of such arrangements is P77 · 8 · 7 · 6 · 5 · 4 = 7!8!/3!. 2) Exercise 1, #3 (i) Solution. There are m + n students to arrange. The number of arrangements is m+n Pm+n = (m + n)!. (ii) Solution. We use the following steps to arrange: 1 · Step 1: Arrange the n girls. There are Pnn different ways. · Step 2: Put a boy into one of the n + 1 spaces in G1 . . . Gn . The number of possibilities is n + 1. · Step 3: Put a boy into one of the remaining n spaces. There are n possibilities. .. . · Step m + 1: Put a boy into one of the remaining n − m + 2 spaces. There are n − m + 2 possibilities. By the Multiplication Principle, the number of such arrangements is Pnn · (n + 1) · n · · · (n − m + 2) = n!(n + 1)!/(n − m + 1)!. (iii) Solution. We perform the following: · Step 1: Treating the n girls as one block and each boy as one block, arrange the m + 1 m+1 blocks. The number of possibilities is Pm+1 . · Step 2: Arrange the n girls in the block. There are pnn ways. By the Multiplication Principle, the number of such arrangements is m+1 n Pm+1 Pn = (m + 1)!n!. (iv) Solution. The steps for such arrangements are: · Step 1: Treating the boy and the girl as one block and each of the other students as m+n−1 one block, arrange the m + n − 1 blocks. The number of ways is Pm+n−1 . · Step 2: Arrange the boy and the girl within the block. There are P22 possibilities. By the Multiplication Principle, the number of such arrangements is m+n−1 2 Pm+n−1 P2 = (m + n − 1)!2!. 3) Exercise 1, #4 i) Solution. Since the number of such words is exactly the number of 5-permutations of the 10 letter, there are P510 = 10!/5! such 5-letter words. ii) Solution. We form the 5-letter words in these steps: 2 · Step 1: Arrange three of A, B, C, D, E, F and put them into the first, third and fifth position. The number of possibilities is P36 . · Step 2: Put two of G, H, I, J into the second and fourth position. The number of possibilities is P24 . By the Multiplication Principle, the number of such words is P36 P24 = 120 · 12 = 1440. 4) Exercise 1, #5 Solution. We perform the following steps: · Step 1: Form a block with 7 letters, which starts with x, ends with y and has 5 other letters in between. There are P524 possibilities. · Step 2: Treating each of the remaining 19 letters as a single block, arrange the 20 20 blocks. The number of possibilities is P20 . 20 By the Multiplication Principle, the number of such words is P524 P20 = 24!20!/5!. 5) Exercise 1, #10 Solution. We have the following prime factorizations: 1040 = 240 · 540 , 2030 = 260 · 530 . The natural number d is a common divisor if and only if d = 2α 5β for integers α, β with 0 ≤ α ≤ 40 and 0 ≤ β ≤ 30. Since any common divisor is determined by first picking a (with 41 possibilities) and then picking b (with 31 possibilities). By the Multiplication Principle, the number of common divisors is 41 · 31 = 2171. 6) Exercise 1, #11 (ii) (ii) Solution. Since n = 2 · 32 · 5 · 7, any divisor of n that 3 divides is of the form d = 2α 3β 5γ 7δ , where α, β, γ, δ are integers with 0 ≤ α ≤ 1, 1 ≤ β ≤ 2, 0 ≤ γ ≤ 1, 0 ≤ δ ≤ 1. Because any such d is determined by picking α, β, γ, δ in a sequence of steps, the Multiplication Principle implies that the number of divisors of n which are multiples of 3 is 2 · 2 · 2 · 2 = 16. 3 7) Exercise 1, #12 (ii) Proof. Suppose n has prime factors p1 , . . . , pm with a prime factorization n = pα1 1 · · · pαmm , where pi > 1 and αi ∈ N for all i. Then 1 m · · · p2α n2 = (pα1 1 · · · pαmm )2 = p2α 1 m . Any positive divisor of n2 is of the form d = pβ1 1 · · · pβmm , where βi is an integer with 0 ≤ βi ≤ 2αi for each i. The total number of such divisors, by the Multiplication Principle, is k = (2α1 + 1)(2α2 + 1) · · · (2αm + 1). Since each of the finitely many 2αi + 1 is odd and the product of two odd numbers is odd, by induction we see that k is odd. Therefore, n2 has an odd number of positive divisors. 4