Difference between revisions of "15-344/Homework Assignment 7"

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Homework Assignment 8 Solution
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'''15 How many 8-digit sequences are there involving exactly six different digits?'''
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'''A)'''
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We have 2 cases: 1 digit repeated three times (for example, 11123456) and 2 digits repeated (for example, 11223456)
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For the first case, there are <math> \binom{10}{6} </math> ways to choose 6 digits from 10 digit choice. Then there are
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<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>
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Similarly, for the second case, we have: <math>\binom{10}{6} \binom{6}{2}\frac{8!}{2!2!}</math>
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The sum of the two cases gives the answer:
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<math>\binom{10}{6}(\binom{6}{1}\frac{8!}{3!}+ \binom{6}{2}\frac{8!}{2!2!})</math>

Revision as of 21:36, 15 December 2015

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 8 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!})