15-344/Homework Assignment 7

DBN: Classes: 2015-16: 15-344 / Navigation Welcome to Math 344! Edits to the Math 344 web sites no longer count for the purpose of good deed points. # Week of... Notes and Links 1 Sep 14 About This Class, Day One Handout, Tuesday, Hour 3 Handout, Thursday, Tutorial 1 Handout 2 Sep 21 Tuesday, Tutorial 2 Page 1, Tutorial 2 Page 2, Thursday, HW1 3 Sep 28 Tuesday, Class Photo, Tutorial 3 Page 1, Tutorial 3 Page 2, Thursday, HW2 4 Oct 5 Tuesday, Drawing $n$-cubes, Thursday, HW3 5 Oct 12 Tuesday, Tutorial Handout, Thursday, HW4 6 Oct 19 Tuesday, Tutorial Handout, Thursday 7 Oct 26 Term Test on Tuesday, Dijkstra Handout,Thursday,HW5 8 Nov 2 Tuesday, Tutorial Handout, Thursday, HW6, Sunday November 8 is the last day to drop this class 9 Nov 9 Monday-Tuesday is UofT Fall Break, Thursday, HW7 10 Nov 16 Tuesday, Thursday, HW8 11 Nov 23 Tuesday, Thursday, HW9 12 Nov 30 Tuesday, Thursday, HW10 13 Dec 7 Tuesday, FibonacciFormula.pdf, semester ends on Wednesday - no class Thursday F Dec 11-22 The Final Exam Register of Good Deeds Add your name / see who's in!

This assignment is due at the tutorials on Thursday November 19. Here and everywhere, neatness counts!! You may be brilliant and you may mean just the right things, but if the teaching assistants will be having hard time deciphering your work they will give up and assume it is wrong.

Reread sections 5.1-5.4 of our textbook. Remember that reading math isn't like reading a novel! If you read a novel and miss a few details most likely you'll still understand the novel. But if you miss a few details in a math text, often you'll miss everything that follows. So reading math takes reading and rereading and rerereading and a lot of thought about what you've read. Also, preread sections 5.5 and 6.1, just to get a feel for the future.

Solve problems 8, 10, 15, 20, 24, 28, and 29 in section 5.3 and problems 1, 3, 16, 39, 40, 63, and 65 in section 5.4, but submit only your solutions of the underlined problems.

Dror's notes above / Students' notes below

Homework Assignment 7 Solution

15 How many 8-digit sequences are there involving exactly six different digits?

A) We have 2 cases: 1 digit repeated three times (for example, 11123456) and 2 digits repeated (for example, 11223456)

For the first case, there are ${\binom {10}{6}}$ ways to choose 6 digits from 10 digit choice. Then there are ${\frac {8!}{3!}}$ arrangements involving a triple digit and and ${\binom {6}{1}}$ different triple digits possible (111,222 and so on).So, in total, we have: ${\binom {10}{6}}{\binom {6}{1}}{\frac {8!}{3!}}$

Similarly, for the second case, we have: ${\binom {10}{6}}{\binom {6}{2}}{\frac {8!}{2!2!}}$

The sum of the two cases gives the answer:

${\binom {10}{6}}({\binom {6}{1}}{\frac {8!}{3!}}+{\binom {6}{2}}{\frac {8!}{2!2!}})$

24 How many arrangements of the word PREPOSTEROUS are there in which the five vowels are consecutive?

A)

We have P-2, R-2, E-2, O-2, S-2, T-1,U-1

The vowels are OOUEE and its variations.

The number of all the possible variations of OOUEE is: ${\frac {5!}{2!2!}}$

The remaining 7 letters can be arranged in ${\frac {7!}{2!2!2!}}$ ways.

Also, there are 8 different positions for the vowel group in the word. For example, OOUEE_ _ _ _ _ _ _ and _ _ _ _ _ _ _ OOUEE

In total, $8{\frac {5!}{4}}{\frac {7!}{8}}=151200$

28 How many ways are there to place nine different rings on the four fingers of your right hand if a) the order of the rings on a finger does not matter b) the order of the rings on a finger is considered

A) a) For each ring, we have a choice of being on 1 - 4th finger. So, we have $4^{9}$ possible such ways.

A) b) First, without considering the different types of the rings, consider the distribution of the nine rings to four fingers: ${\binom {n+k-1}{k-1}}={\binom {9+4-1}{4-1}}={\binom {12}{3}}$ If we consider the fact that the rings are all different, we have the permutations $9!$ So, in total, $9!{\binom {12}{3}}$ ways.

39 How many non-negative integer solutions are there to the equation $2x_{1}+2x_{2}+x_{3}+x_{4}=12$.

A) This can be viewed as a special type of distribution problems.

First we think of all the possible pairs of $x_{1},x_{2}$. For example:

$x_{1}=0,x_{2}=0\to x_{3}+x_{4}=12$

$x_{1}=1,x_{2}=0\to x_{3}+x_{4}=10$ (two such cases since we can interchange the role of x1 and x2)

and so on.

In total, we have:

${\binom {12+2-1}{2-1}}+2{\binom {10+2-1}{2-1}}+3{\binom {8+2-1}{2-1}}...+7{\binom {0+2-1}{2-1}}=140ways$

Scanned Solution to Homework7 P29 in 5.3, P3,39,63 in 5.4

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