15-344/Homework Assignment 7 - Revision history

Octopusheng at 06:48, 16 December 2015

2015-12-16T06:48:28Z

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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>		<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>

			== Scanned Solution to Homework7 P29 in 5.3, P3,39,63 in 5.4 ==
			<gallery>
			15-344-HW7-Yunheng Chen.jpg\|Page 1
			</gallery>

2015-12-16T02:22:23Z

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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)		<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)

			and so on.

	In total, we have:		In total, we have:

2015-12-16T02:21:58Z

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

	First we think of all the possible pairs of <math>x_{1}, x_{2} </math>. For example:		First we think of all the possible pairs of <math>x_{1}, x_{2} </math>. For example:

	<math>x_{1} = 0, x_{2} = 0 \to x_{3}+x_{4} = 12 </math>		<math>x_{1} = 0, x_{2} = 0 \to x_{3}+x_{4} = 12 </math>

	<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)		<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)

2015-12-16T02:21:31Z

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	<math> \binom{n+k-1}{k-1} = \binom{9+4-1}{4-1} = \binom{12}{3} </math>		<math> \binom{n+k-1}{k-1} = \binom{9+4-1}{4-1} = \binom{12}{3} </math>
	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.		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.

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

			'''A)''' This can be viewed as a special type of distribution problems.
			First we think of all the possible pairs of <math>x_{1}, x_{2} </math>. For example:

			<math>x_{1} = 0, x_{2} = 0 \to x_{3}+x_{4} = 12 </math>
			<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)

			In total, we have:

			<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>

2015-12-16T02:11:18Z

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	{{15-344:Dror/Students Divider}}		{{15-344:Dror/Students Divider}}

	Homework Assignment 8 Solution		Homework Assignment 7 Solution

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

2015-12-16T02:01:04Z

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	In total, <math>8\frac{5!}{4}\frac{7!}{8} = 151200 </math>		In total, <math>8\frac{5!}{4}\frac{7!}{8} = 151200 </math>

			'''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 <math> 4^9 </math> possible such ways.

			'''A) b)''' First, without considering the different types of the rings, consider the distribution of the nine rings to four fingers:
			<math> \binom{n+k-1}{k-1} = \binom{9+4-1}{4-1} = \binom{12}{3} </math>
			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.

2015-12-16T01:47:59Z

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	The vowels are OOUEE and its variations.		The vowels are OOUEE and its variations.

	The number of all the possible variations of OOUEE is: <math>\frac{5!}/{2!2!}</math>		The number of all the possible variations of OOUEE is: <math>\frac{5!}{2!2!}</math>

	The remaining 7 letters can be arranged in <math>\frac{7!}{2!2!2!}</math> ways.		The remaining 7 letters can be arranged in <math>\frac{7!}{2!2!2!}</math> ways.

2015-12-16T01:47:34Z

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	<math>\binom{10}{6}(\binom{6}{1}\frac{8!}{3!}+ \binom{6}{2}\frac{8!}{2!2!})</math>		<math>\binom{10}{6}(\binom{6}{1}\frac{8!}{3!}+ \binom{6}{2}\frac{8!}{2!2!})</math>

			'''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: <math>\frac{5!}/{2!2!}</math>

			The remaining 7 letters can be arranged in <math>\frac{7!}{2!2!2!}</math> ways.

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

			In total, <math>8\frac{5!}{4}\frac{7!}{8} = 151200 </math>

2015-12-16T01:36:57Z

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	{{15-344:Dror/Students Divider}}		{{15-344:Dror/Students Divider}}

			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 <math> \binom{10}{6} </math> ways to choose 6 digits from 10 digit choice. Then there are
			<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>

			Similarly, for the second case, we have: <math>\binom{10}{6} \binom{6}{2}\frac{8!}{2!2!}</math>

			The sum of the two cases gives the answer:

			<math>\binom{10}{6}(\binom{6}{1}\frac{8!}{3!}+ \binom{6}{2}\frac{8!}{2!2!})</math>

2015-12-16T00:39:30Z

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	{{15-344:Dror/Students Divider}}		{{15-344:Dror/Students Divider}}

	In problem 29, Section 5.3, we were asked to prove that

	<math>\sum_{k_1 + k_2 + k_3 = 10} P(10, k_1, k_2, k_3) = 3^{10}</math>, where <math>k_1, k_2, k_3 \in \mathbb{Z}_{\geq 0}</math>.

	It is tempting to think that this is a special case of some identity and that a general pattern can be found.

	Replacing <math>3</math> with any <math>m \in \mathbb{Z}_{>0}</math> and <math>10</math> with any <math>n \in \mathbb{Z}_{\geq0}</math> gives

	<math>\sum_{k_1+k_2+\cdots+k_m=n} P(n, k_1, k_2, \ldots, k_m) = m^n</math>

	Does it hold, or can it be generalized to a greater extent?

	Yes. We have the [https://en.wikipedia.org/wiki/Multinomial_theorem Multinomial_theorem].

	<math>(\sum_{i=1}^m x_i)^n = \sum_{k_1 + k_2 + \cdots + k_m = n} P(n, k_1, k_2, \ldots, k_m) \prod_{j=1}^m x_{j}^{k_{j}}</math>, where <math>m \in \mathbb{Z}_{>0}, n \in \mathbb{Z}_{\geq0}</math> and <math>x_1, \ldots, x_m\in \mathbb{R}</math>.

	For a proof by induction, see [https://proofwiki.org/wiki/Multinomial_Theorem here].