Difference between revisions of "15-344/Homework Assignment 8"

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, 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
 
, 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
 
<math> n^2 </math>. Their sum gives the total number of possibilities, which is exactly the right hand side of the equality.
 
<math> n^2 </math>. Their sum gives the total number of possibilities, which is exactly the right hand side of the equality.
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== Scanned Solution to Homework8 P7,11,15 ==
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15-344-HW8-Yunheng Chen.jpg|Page 1
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Revision as of 01:54, 16 December 2015

This assignment is due at the tutorials on Thursday November 26. 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 5.1-5.5 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.

Solve and submit problems 7, 9a, 11, and 15 in section 5.5. (9b was originally also assigned, but it is too difficult).

Dror's notes above / Students' notes below

Homework Assignment 8 Solution

9a) Show by combinatorial argument that \binom{2n}{n} = 2 \binom{n}{2} + n^2

A)Split  2n into two groups of n. Then choosing 2 out of  2n can be seen as either choosing 2 from group 1 only , which is \binom{n}{2} or 2 from group 2 only, which is also \binom{n}{2} , or one from each group, which is  n^2 . Their sum gives the total number of possibilities, which is exactly the right hand side of the equality.

Scanned Solution to Homework8 P7,11,15