AMS 301 - Homework 8 Solutions

8.1.10

If we have A be the set of arrangements that end with an 8 and B be the set of arrangements thatbegin with a 3, we are looking for N(A∩B). By Inclusion/Exclusion, this is equal to N −N(A)−N(B) + N(A ∩ B), which works out to 10! − 9! − 9! + 8! = 10! − 2 · 9! + 8!.

8.1.12

a. If A is the set of sequences that contain THE and B is the set of sequences that contain AID,we are looking for N(A ∪ B) = N(A) + N(B) − N(A ∩ B). Gluing letters together, tell usthat N(A) = N(B) = 24! and N(A ∩ B) = 22!. Therefore, the answer is 2 · 24! − 22!.

b. If A is the set of sequences that contain THE and B is the set of sequences that containMATH, we are looking for N(A ∩ B) = N − N(A) − N(B) + N(A ∩ B). Gluing letterstogether (note that the sequences are not independent, THE and MATH share letters), weget that this is equal to 26! − 24! − 23! + 22!.

8.1.18

We have that 280 = 23·5 ·7, so a number is relatively prime to 280 if and only if it is not divisible by

2, 5, or 7. That is, if A is the numbers divisible by 2, B is the numbers divisible by 5, and C is thenumbers divisible by 7, we are looking for N(A∩B ∩C). Using inclusion/exclusion, this works outto N−N(A)−N(B)−N(C)+N(A∩B)+N(A∩C)+N(B∩C)−N(A∩B∩C). Working this out, weget 280− 280

2−

280

5−

280

7+ 280

2·5+ 280

2·7+ 280

5·7−

280

2·5·7, which yields 280−140−56−40+28+20+8−4 = 96.

8.2.2

Let A1 be the set of rolls that contain a 1, A2 be the rolls that contain a 2, and so on. Then, thequestion translates to finding N(A1 ∩A2 ∩A3 ∩A4 ∩A5 ∩A6). By inclusion/exclusion, this worksout to N −N(A1)−N(A2)− . . . (this sum has 26 = 64 terms, so don’t try to write them all out).To figure out the value, we use symmetry. Note that because the numbers in each of these setsdon’t change if we permute the values on the dice, there is no difference between N(A1) and N(A2),and similarly between all of the singles, doubles, triples, quadruples, and quintuples, respectively.That is, because of symmetry, this sum of 64 terms reduces to N −C(6, 1)N(A1) + C(6, 2)N(A1 ∩

A2)−C(6, 3)N(A1 ∩A2 ∩A3)+C(6, 4)N(A1 ∩A2 ∩A3 ∩A4)−C(6, 5)N(A1 ∩A2 ∩A3 ∩A4 ∩A5)+C(6, 6)N(A1 ∩ A2 ∩ A3 ∩ A4 ∩ A5 ∩ A6). Now, we know that N(A1) = 59 because it is the numberof possible outcomes that do not contain a 1. Similarly, N(A1∩A2) = 49, and so on. Therefore, wehave that the answer is 69

− C(6, 1)59 + C(6, 2)49− C(6, 3)39 + C(6, 4)29

− C(6, 5)19 + C(6, 6)09.

8.2.30

Let A be the set of arrangements that have adjacent a’s, B be the arrangements that have adjacentb’s and C be the arrangements that have adjacent c’s. Then, we’re looking for N(A∩B ∩C). Thisis a standard inclusion/exclusion, so we’re looking for N − N(A) − N(B) − N(C) + N(A ∩ B) +N(A∩C)+N(B∩C)−N(A∩B∩C). Using symmetry, this breaks down to N −3N(A)+3N(A∩

B) − N(A ∩ B ∩ C).To figure out N(A), we try to use the standard ’gluing’ idea. Take a pair of a’s and glue them

together, then arrange the letters a, aa, b, b, b, c, c, c in 8!

1!1!3!3!ways. Note that every arrangement

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that contains adjacent a’s is found by this approach, but there are some that are counted twice. Forexample a, aa, b, b, b, c, c, c and aa, a, b, b, b, c, c, c are considered different arrangements according tothis gluing approach. This bad situation occurs exactly when aaa appears in the arrangement, soto fix our count, we just need to subtract the triples. Therefore, N(A) = 8!

1!1!3!3!−

7!

1!3!3!.

Now for N(A ∩ B). In a similar vein to N(A), we glue a pair of a’s and a pair of b togetherand then arrange the remaining letters in 7!

1!1!1!1!3!ways. Just like before though, we’ve counted too

many arrangements. The arrangements we overcounted are exactly the ones with triple a’s or withtriple b’s. Using inclusion/exclusion on this tells us that we overcounted by 6!

1!1!1!3!+ 6!

1!1!1!3!−

5!

1!1!3!,

meaning that N(A ∩ B) = 7!

1!1!1!1!3!− 2 ·

6!

1!1!1!3!+ 5!

1!1!3!.

The last term N(A ∩B ∩C) is computed identically, but with a three term inclusion/exclusioncalculation. That is, glue a pair of a’s, b’s, and c’s together, respectively, and then arrange theletters in 6!

1!1!1!1!1!1!ways. Our overcounting is exactly the arrangements with triple a’s, triple

b’s, or triple c’s. This means we overcount by exactly 3 ·5!

1!1!1!1!1!− 3 ·

4!

1!1!1!1!+ 3!

1!1!1!. Therefore,

N(A ∩ B ∩ C) = 6!

1!1!1!1!1!1!− 3 ·

5!

1!1!1!1!1!+ 3 ·

4!

1!1!1!1!−

3!

1!1!1!.

Putting it all together, we get that the number of arrangement with no pair of adjacent a’s, b’s,or c’s, is

N(A ∩ B ∩ C) =9!

3!3!3!− 3 · (

8!

1!1!3!3!−

7!

1!3!3!)

+ 3 · (7!

1!1!1!1!3!− 2 ·

6!

1!1!1!3!+

5!

1!1!3!)

− (6!

1!1!1!1!1!1!− 3 ·

5!

1!1!1!1!1!+ 3 ·

4!

1!1!1!1!−

3!

1!1!1!)

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