http://drorbn.net/api.php?action=feedcontributions&user=Octopusheng&feedformat=atomDrorbn - User contributions [en]2024-03-29T10:03:49ZUser contributionsMediaWiki 1.21.1http://drorbn.net/index.php?title=15-344/Homework_Assignment_715-344/Homework Assignment 72015-12-16T06:48:28Z<p>Octopusheng: </p>
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This assignment is due <span style="color: blue;">at the tutorials on Thursday November 19</span>. 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.<br />
<br />
'''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.<br />
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'''Solve''' problems 8, 10, <u>15</u>, 20, <u>24</u>, <u>28</u>, and <u>29</u> in section 5.3 and problems 1, <u>3</u>, 16, <u>39</u>, 40, <u>63</u>, and <u>65</u> in section 5.4, but submit only your solutions of the underlined problems.<br />
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{{15-344:Dror/Students Divider}}<br />
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Homework Assignment 7 Solution<br />
<br />
'''15 How many 8-digit sequences are there involving exactly six different digits?'''<br />
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'''A)'''<br />
We have 2 cases: 1 digit repeated three times (for example, 11123456) and 2 digits repeated (for example, 11223456)<br />
<br />
For the first case, there are <math> \binom{10}{6} </math> ways to choose 6 digits from 10 digit choice. Then there are <br />
<math>\frac{8!}{3!}</math> arrangements involving a triple digit and and <math>\binom{6}{1}</math> different triple digits possible (111,222 and so on).So, in total, we have: <math>\binom{10}{6} \binom{6}{1}\frac{8!}{3!}</math><br />
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Similarly, for the second case, we have: <math>\binom{10}{6} \binom{6}{2}\frac{8!}{2!2!}</math><br />
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The sum of the two cases gives the answer:<br />
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<math>\binom{10}{6}(\binom{6}{1}\frac{8!}{3!}+ \binom{6}{2}\frac{8!}{2!2!})</math><br />
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'''24 How many arrangements of the word PREPOSTEROUS are there in which the five vowels are consecutive? '''<br />
<br />
'''A)''' <br />
<br />
We have P-2, R-2, E-2, O-2, S-2, T-1,U-1 <br />
<br />
The vowels are OOUEE and its variations. <br />
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The number of all the possible variations of OOUEE is: <math>\frac{5!}{2!2!}</math> <br />
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The remaining 7 letters can be arranged in <math>\frac{7!}{2!2!2!}</math> ways.<br />
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Also, there are 8 different positions for the vowel group in the word. For example, OOUEE_ _ _ _ _ _ _ and _ _ _ _ _ _ _ OOUEE<br />
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In total, <math>8\frac{5!}{4}\frac{7!}{8} = 151200 </math><br />
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'''28 How many ways are there to place nine different rings on the four fingers of your right hand if <br />
a) the order of the rings on a finger does not matter<br />
b) the order of the rings on a finger is considered'''<br />
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'''A) a)''' <br />
For each ring, we have a choice of being on 1 - 4th finger. So, we have <math> 4^9 </math> possible such ways. <br />
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'''A) b)''' First, without considering the different types of the rings, consider the distribution of the nine rings to four fingers: <br />
<math> \binom{n+k-1}{k-1} = \binom{9+4-1}{4-1} = \binom{12}{3} </math> <br />
If we consider the fact that the rings are all different, we have the permutations <math>9!</math> So, in total, <math>9!\binom{12}{3}</math> ways.<br />
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'''39 How many non-negative integer solutions are there to the equation <math> 2x_{1} + 2x_{2} + x_{3} + x_{4} = 12 </math>.'''<br />
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'''A)''' This can be viewed as a special type of distribution problems. <br />
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First we think of all the possible pairs of <math>x_{1}, x_{2} </math>. For example:<br />
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<math>x_{1} = 0, x_{2} = 0 \to x_{3}+x_{4} = 12 </math><br />
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<math>x_{1} = 1, x_{2} = 0 \to x_{3}+x_{4} = 10 </math> (two such cases since we can interchange the role of x1 and x2)<br />
<br />
and so on.<br />
<br />
In total, we have:<br />
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<math>\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} = 140 ways</math><br />
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== Scanned Solution to Homework7 P29 in 5.3, P3,39,63 in 5.4 == <br />
<gallery><br />
15-344-HW7-Yunheng Chen.jpg|Page 1<br />
</gallery></div>Octopushenghttp://drorbn.net/index.php?title=File:15-344-HW7-Yunheng_Chen.jpgFile:15-344-HW7-Yunheng Chen.jpg2015-12-16T06:47:23Z<p>Octopusheng: </p>
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<div></div>Octopushenghttp://drorbn.net/index.php?title=15-344/Homework_Assignment_615-344/Homework Assignment 62015-12-16T06:47:00Z<p>Octopusheng: /* Scanned Solution to Homework6 P85 in 5.2, P5 in 5.3 */</p>
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<div>{{15-344/Navigation}}<br />
This assignment is due <span style="color: blue;">at the tutorials on Thursday November 12</span>. 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.<br />
<br />
'''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.<br />
<br />
'''Solve''' problems 4, 8, <u>18</u>, <u>39</u>, <u>46</u>, and 49 in section 5.1, problems 4, 9, 16, 23, <u>31</u>, 38, <u>56</u>, 68, <u>72</u>, and <u>85</u> in section 5.2, and problems <u>5</u>, and 29 in section 5.3, but submit only your solutions of the underlined problems.<br />
<br />
Sorry for again posting an assignment a bit late.<br />
<br />
----<br />
<br />
An unrelated matter - I was asked to distribute the following, and I hereby comply:<br />
<br />
Subject: COMC Markers Needed<br />
<br />
We need people to help us mark the 2015 Canadian Open Mathematics Contest. Anyone can help mark (including non-math students so please feel free to pass this message onto your friends and invite them to help too).<br />
<br />
We are having a marking party on November 21 and 22 complete with pizza, pop and other snacks. However we need markers to help the week before and after this as well. Anyone who assists will gain hours towards CCR recognition and certificates are provided to those that wish them.<br />
<br />
If you are able to help please send an email to outreach@math.toronto.edu with "COMC Marker" as the subject and we'll add your name to the list. We're also happy to answer any questions you may have.<br />
<br />
Sincerely,<br />
<br />
Pamela Brittain<br />
<br/>Outreach and Special Projects Officer<br />
<br/>Department of Mathematics<br />
<br/>University of Toronto<br />
<br />
{{15-344:Dror/Students Divider}}<br />
<br />
A solution by student [[Media:15-344-Hw6.pdf|HW6_Solution]]<br />
<br />
== Scanned Solution to Homework6 P85 in 5.2, P5 in 5.3 == <br />
<gallery><br />
15-344-HW6-Yunheng Chen.jpg|Page 1<br />
</gallery></div>Octopushenghttp://drorbn.net/index.php?title=15-344/Homework_Assignment_615-344/Homework Assignment 62015-12-16T06:46:27Z<p>Octopusheng: </p>
<hr />
<div>{{15-344/Navigation}}<br />
This assignment is due <span style="color: blue;">at the tutorials on Thursday November 12</span>. 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.<br />
<br />
'''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.<br />
<br />
'''Solve''' problems 4, 8, <u>18</u>, <u>39</u>, <u>46</u>, and 49 in section 5.1, problems 4, 9, 16, 23, <u>31</u>, 38, <u>56</u>, 68, <u>72</u>, and <u>85</u> in section 5.2, and problems <u>5</u>, and 29 in section 5.3, but submit only your solutions of the underlined problems.<br />
<br />
Sorry for again posting an assignment a bit late.<br />
<br />
----<br />
<br />
An unrelated matter - I was asked to distribute the following, and I hereby comply:<br />
<br />
Subject: COMC Markers Needed<br />
<br />
We need people to help us mark the 2015 Canadian Open Mathematics Contest. Anyone can help mark (including non-math students so please feel free to pass this message onto your friends and invite them to help too).<br />
<br />
We are having a marking party on November 21 and 22 complete with pizza, pop and other snacks. However we need markers to help the week before and after this as well. Anyone who assists will gain hours towards CCR recognition and certificates are provided to those that wish them.<br />
<br />
If you are able to help please send an email to outreach@math.toronto.edu with "COMC Marker" as the subject and we'll add your name to the list. We're also happy to answer any questions you may have.<br />
<br />
Sincerely,<br />
<br />
Pamela Brittain<br />
<br/>Outreach and Special Projects Officer<br />
<br/>Department of Mathematics<br />
<br/>University of Toronto<br />
<br />
{{15-344:Dror/Students Divider}}<br />
<br />
A solution by student [[Media:15-344-Hw6.pdf|HW6_Solution]]<br />
<br />
== Scanned Solution to Homework6 P85 in 5.2, P5 in 5.3 == <br />
<gallery><br />
15-344-HW6-Yunheng Chen|Page 1<br />
</gallery></div>Octopushenghttp://drorbn.net/index.php?title=File:15-344-HW6-Yunheng_Chen.jpgFile:15-344-HW6-Yunheng Chen.jpg2015-12-16T06:45:56Z<p>Octopusheng: </p>
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<div></div>Octopushenghttp://drorbn.net/index.php?title=15-344/Homework_Assignment_115-344/Homework Assignment 12015-12-16T06:22:05Z<p>Octopusheng: </p>
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<div>{{15-344/Navigation}}<br />
This assignment is due <span style="color: blue;">at the tutorials on Thursday October 1</span>. 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.<br />
<br />
'''Reread''' Chapter 1 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''' chapter 2, just to get a feel for the future.<br />
<br />
'''Solve''' problems 4 and 16 for section 1.1, problems 3, 6, and 11 for section 1.2, and problems 1, 2, and 12 for section 1.3.<br />
<br />
{{15-344:Dror/Students Divider}}<br />
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== Scanned Solution to Homework1 == <br />
<gallery><br />
15-344-HW1.1-Yunheng Chen.jpg|Page 1<br />
15-344-HW1.2-Yunheng Chen.jpg|Page 2<br />
15-344-HW1.3-Yunheng_Chen.jpg|Page 3<br />
15-344-HW1.4-Yunheng_Chen.jpg|Page 4<br />
15-344-HW1.5-Yunheng_Chen.jpg|Page 5<br />
</gallery></div>Octopushenghttp://drorbn.net/index.php?title=File:15-344-HW1.5-Yunheng_Chen.jpgFile:15-344-HW1.5-Yunheng Chen.jpg2015-12-16T06:21:41Z<p>Octopusheng: </p>
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<div></div>Octopushenghttp://drorbn.net/index.php?title=File:15-344-HW1.4-Yunheng_Chen.jpgFile:15-344-HW1.4-Yunheng Chen.jpg2015-12-16T06:21:21Z<p>Octopusheng: </p>
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<div></div>Octopushenghttp://drorbn.net/index.php?title=File:15-344-HW1.3-Yunheng_Chen.jpgFile:15-344-HW1.3-Yunheng Chen.jpg2015-12-16T06:21:07Z<p>Octopusheng: </p>
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<div></div>Octopushenghttp://drorbn.net/index.php?title=File:15-344-HW1.2-Yunheng_Chen.jpgFile:15-344-HW1.2-Yunheng Chen.jpg2015-12-16T06:20:44Z<p>Octopusheng: </p>
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<div></div>Octopushenghttp://drorbn.net/index.php?title=File:15-344-HW1.1-Yunheng_Chen.jpgFile:15-344-HW1.1-Yunheng Chen.jpg2015-12-16T06:20:19Z<p>Octopusheng: </p>
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<div></div>Octopushenghttp://drorbn.net/index.php?title=15-344/Homework_Assignment_915-344/Homework Assignment 92015-12-16T05:58:24Z<p>Octopusheng: </p>
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<div>{{15-344/Navigation}}<br />
This assignment is due <span style="color: blue;">at the tutorials on Thursday December 3</span>. 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.<br />
<br />
'''Reread''' sections 6.1-6.2 of our textbook, '''and''' your notes for November 26 and December 1. 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''' chapter 7, just to get a feel for the future.<br />
<br />
'''Solve''' problems 2abde, <u>2c</u>, 4abc, <u>4d</u>, 12, <u>16</u>, 20, and 25 in section 6.1 and problems 1, <u>2</u>, 11abcd, <u>11e</u>, 30, <u>32</u>, and <u>43</u> in section 6.2, but submit only your solutions of the underlined problems.<br />
<br />
'''In addition,''' solve the following problem. There is no need to submit your solution.<br />
<br />
'''Part 1.''' For <math>n\geq 0</math>, let <math>C_n</math> be the <math>n</math>'th Catalan number, defined as the number of words made of <math>n</math> <math>a</math>'s and <math>n</math> <math>b</math>'s in which in every initial segment there are at least as many <math>a</math>'s as <math>b</math>'s. Prove that for <math>n\geq 1</math>, <math>C_n=C_0C_{n-1}+C_1C_{n-2}+C_2C_{n-3}+\ldots+C_{n-1}C_0</math>.<br />
<br />
''Hint.'' If <math>w</math> is such a word, let <math>k\geq 1</math> be the smallest such that if you cut <math>w</math> after <math>2k</math> letters, then the number of <math>a</math>'s before the cut is equal to the number of <math>b</math>'s before the cut. What can you say about the part of <math>w</math> before the cut, and the part of <math>w</math> after the cut?<br />
<br />
'''Part 2.''' For <math>n\geq 1</math>, let <math>D_n</math> be the number of different ways of computing the product <math>A_1A_2\cdots A_n</math> of <math>n</math> matrices. Prove that for <math>n\geq 2</math>, <math>D_n=D_1D_{n-1}+D_2D_{n-2}+D_3D_{n-3}+\ldots+D_{n-1}D_1</math>.<br />
<br />
''Hint.'' In the last matrix multiplication to be carried out, you'd be multiplying the product of <math>k</math> of the matrices with the product of <math>n-k</math> of the matrices.<br />
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'''Part 3.''' Let <math>E_2=1</math> and for <math>n\geq 3</math> let <math>E_n</math> be the number of triangulations of a convex <math>n</math>-gon using non-crossing diagonals. Prove that for <math>n\geq 3</math>, <math>E_n=E_2E_{n-1}+E_3E_{n-2}+E_4E_{n-3}+\ldots+E_{n-1}E_2</math>.<br />
<br />
''Hint.'' Choose one edge of the <math>n</math>-gon, and consider the triangle on top of it, what's to its left, and what's to its right.<br />
<br />
'''Part 4.''' Use the above three assertions to prove that for any <math>n\geq 0</math>, <math>C_n=D_{n+1}=E_{n+2}</math>.<br />
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{{15-344:Dror/Students Divider}}<br />
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Homework Assignment 9 Solution<br />
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'''2) Build a generating function for <math>a_{r}</math>, the number of integer solutions to the following equations.'''<br />
<br />
c) <math>e_{1}+e_{2}+e_{3}+e_{4} = r, 2\leq e_{i} \leq 7, e_{1}</math> even <math> e_{2} </math> odd.<br />
<br />
'''A)''' For <math>e_{3}</math> and <math>e_{4}</math> together, we have <math>g_{1}(x) = (x^2+x^3+x^4+x^5+x^6+x^7)^2 </math><br />
<br />
For <math>e_{1}</math>, we have <math>g_{2}(x) = (x^2+x^4+x^6) </math><br />
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For <math>e_{2}</math>, we have <math>g_{3}(x) = (x^3+x^5+x^7) </math><br />
<br />
In total, we have <math>g(x) = (x^2+x^4+x^6)(x^3+x^5+x^7)(x^2+x^3+x^4+x^5+x^6+x^7)^2 </math><br />
<br />
'''4 d) Three different boxes with at most five objects in the first box'''<br />
<br />
'''A)''' <math>e_{1}+e_{2}+e_{3} = r</math> with <math>0\leq e_{1} \leq 5</math><br />
<br />
For <math>e_{1}</math>, we have <math> (1+x+x^2+x^3+x^4+x^5) </math> <br />
<br />
Note that for the other two, we don't have a limit, so:<br />
<br />
<math>\sum_{k=0}^{\infty}x^{k} = \frac{1}{1-x} </math><br />
<br />
so, in total we have: <math> g(x) = (1+x+x^2+x^3+x^4+x^5)(\frac{1}{1-x})^2 </math><br />
<br />
<br />
'''16) Find a generating function for the number of integers between 0 and 999,999 whose sum of digits is <math> r </math>.'''<br />
<br />
'''A)''' For each integer in the range, we can express them as a six digit number, including 0's in the front (5 will be 000005). <br />
So this amounts to finding:<br />
<br />
<math> e_{1}+e_{2}+e_{3}+e_{4}+e_{5}+e_{6} = r , 0 \leq e_{i} \leq 9 , 1 \leq i \leq 6 </math><br />
<br />
So, we have <math> g(x) = (1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9)^6 </math><br />
<br />
<br />
'''Section 6.1 2) Find the coefficient of <math> x^r </math> in <math> (x^5 + x^6 + x^7 +...)^8</math>'''<br />
<br />
'''A)'''<br />
<br />
<math>(x^5 + x^6 + x^7 +...)^8 = (x^5(1+ x + x^2 ...))^8 = x^{40} \frac{1}{(1-x)^8} </math><br />
<br />
The coefficient of <math> x^r </math> in this will be of <math> x^{r-40} </math> in <math> (1-x)^{-8} </math>, which is<br />
<br />
<math>\binom{(r-40) + 8 - 1}{r-40} </math><br />
<br />
== Scanned Solution to Homework9 P11e,32,43 in Secton6.2 == <br />
<gallery><br />
15-344-HW9.1-Yunheng Chen.jpg|Page 1<br />
HW9.2-Yunheng Chen.jpg|Page 2<br />
</gallery></div>Octopushenghttp://drorbn.net/index.php?title=File:15-344-HW9.1-Yunheng_Chen.jpgFile:15-344-HW9.1-Yunheng Chen.jpg2015-12-16T05:55:52Z<p>Octopusheng: </p>
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<div></div>Octopushenghttp://drorbn.net/index.php?title=File:HW9.2-Yunheng_Chen.jpgFile:HW9.2-Yunheng Chen.jpg2015-12-16T05:55:17Z<p>Octopusheng: </p>
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<div></div>Octopushenghttp://drorbn.net/index.php?title=15-344/Homework_Assignment_815-344/Homework Assignment 82015-12-16T05:54:09Z<p>Octopusheng: </p>
<hr />
<div>{{15-344/Navigation}}<br />
This assignment is due <span style="color: blue;">at the tutorials on Thursday November 26</span>. 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.<br />
<br />
'''Reread''' sections 5.1-5.5 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 6.1-6.3, just to get a feel for the future.<br />
<br />
'''Solve and submit''' problems 7, 9a, 11, and 15 in section 5.5. (9b was originally also assigned, but it is too difficult).<br />
<br />
{{15-344:Dror/Students Divider}}<br />
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Homework Assignment 8 Solution<br />
<br />
'''9a) Show by combinatorial argument that <math>\binom{2n}{n} = 2 \binom{n}{2} + n^2 </math>'''<br />
<br />
'''A)'''Split <math> 2n </math> into two groups of n. Then choosing 2 out of <math> 2n </math> can be seen as either choosing 2 from group 1 only<br />
, which is <math>\binom{n}{2} </math> or 2 from group 2 only, which is also <math>\binom{n}{2} </math>, or one from each group, which is<br />
<math> n^2 </math>. Their sum gives the total number of possibilities, which is exactly the right hand side of the equality.<br />
<br />
== Scanned Solution to Homework8 P7,11,15 == <br />
<gallery><br />
15-344-HW8-Yunheng Chen.jpg|Page 1<br />
</gallery></div>Octopushenghttp://drorbn.net/index.php?title=File:15-344-HW8-Yunheng_Chen.jpgFile:15-344-HW8-Yunheng Chen.jpg2015-12-16T05:51:59Z<p>Octopusheng: </p>
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<div></div>Octopusheng