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Combinatorics Hw1 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.
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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.
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