Assignment 2

course Mth 152

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Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

query 11.2.12 find 10! / [ 4! (10-4)! ] without calculator

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RESPONSE -->

10! [4! (6)!] 10! / 24!

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15:27:16

** 10! / [ 4! * (10-4) ! ] can be simplified to get 10! / ( 4! * 6! ).

This gives you 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / [ ( 4 * 3 * 2 * 1) * ( 6 * 5 * 4 * 3 * 2 * 1) ] .

The numerator and denominator could be multiplied out but it's easier and more instructive to divide out like terms.

Dividing ( 6 * 5 * 4 * 3 * 2 * 1) in the numerator by ( 6 * 5 * 4 * 3 * 2 * 1) in the denominator leaves 10 * 9 * 8 * 7 / (4 * 3 * 2 * 1).

Every factor of the denominator divides into the numerator without remainder: Divide 4 into 8, divide 3 into 9 and 2 into 10 and you get 5 * 3 * 2 * 7 = 210.

NOTE ON WHAT NOT TO DO:

You could figure out that 10! = 3628800, and that 4! * 6! = 24 * 720 = 16480, then finally divided 3628800 by 16480. But that would process would lose accuracy and be ridiculously long for something like 100 ! / ( 30! * 70!). Much better to simply divide out like factors until the denominator goes away. **

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RESPONSE -->

I did not work the bottom number that way. I wasn't real sure how to do it. But after reading your explanation, I understand now

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15:28:41

query 11.2.25 3 switches in a row; fund count prin to find # of possible settings

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RESPONSE -->

3*2*1 = 6 possible settings.

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15:30:29

** There are two possible settings for the first switch, two for the second, two for the third. The setting of one switch in independent of the setting of any other switch so the fundamental counting principle holds. There are therefore 2 * 2 * 2 = 8 possible setting for the three switches.

COMMON ERROR: There are six possible settings and I used fundamental counting principle : first choice 3 ways, second choice 2 ways and third choice 1 way or 3 times 2 times 1 equals 6 ways.

INSTRUCTOR CRITIQUE: You're choosing states of the switches and there are only two states on each. **

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RESPONSE -->

I did the common error way but I see now that the switch can only be on or off so there is only two choices for switch.

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15:38:02

query 11.2.27 If no two adjacent switches are off why does the fundamental counting principle not apply?

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Because it isn't uniform, the switches would not be consistent for each solution.

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15:38:15

** The reason is that the Fund. Counting Principle requires that the events be independent. Here we have the state of one switch influencing the state of its neighbors (neither neighbor can be the same as that switch). The Fund. Counting Principle requires that the events be independent. **

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15:39:17

query 11.2.36 How many odd 3-digit #'s from the set {3, 4, 5}?

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RESPONSE -->

3*2*1 = 6 sets

1, 3, 5 3 sets

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15:57:39

** Using the box method:

1st can be any of the three so the first number of possibilities is 3

2nd number can also be any of the three so the second number of possibilities is 3

The last digit must be odd, so there are only 2 choices for it.

We therefore have 3*3*2=18 possible combinations.

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RESPONSE -->

I didn't get the same answer. I did the 3*2*1 to get 6 but I see now what I did.

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16:00:42

query 11.2.50 10 guitars, 4 cases, 6 amps, 3 processors; # possible setups

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RESPONSE -->

10*4*6*3 = 720 possible set ups

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16:01:07

** A setup consists of a guitar, a case, an amp and a processor.

There are 10 choices for the guitar, 4 for the case, 6 for the amp and 3 for the processor.

So there are 10 * 4 * 6 * 3 = 720 possible setups. **

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RESPONSE -->

I can't believe it, I got it right!!!!

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16:01:34

Add comments on any surprises or insights you experienced as a result of this assignment.

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Having a hard time grasping the concept.

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These ideas are challenging. You're doing a good job and appear to understand either the problems or the given solutions. Keep at it and hopefully by the end of the chapter everything will have sorted itself out.