#$&*
course MTH 152
Please make note of my comments in bold font, and see my comments at the end.
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution.
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
002. `query 2
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Self-critique Rating:
question 10.2.14 (previuosly 11.2.12) find 10! / [ 4! (10-4)! ] without calculator
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Your Solution:
- I note that the factorial seems to represent the “descending value” I mention in the previous assignment of section 2. And that 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 *1.
- Next I fence with the jumbled up mess that is the denominator, of which I shall name the “dominator denominator.” Which really isn’t that bad, but shall be dealt with thusly:
- Standard mathematical procedure dictates the use of the acronym PEMDAS as a go-to guide to deal with the hairier stuff. In this case, I would simplify the dominator denominator from
- [4! (10 - 4)!] to:
- [4! (6)!] or in a stretched out version, [4 * 3 * 2 * 1 (6* 5 * 4 * 3 * 2 *1)], where I take into account the law of distribution which is made apparent by the parenthetical notations listed around the quantity 10 - 4, of which there is another factorial symbol.
OR
- I could think that, without working the 10 - 4 out yet, that the ! is distributed to both 10 and 4. Hmm. I’ve always been at odds with myself when it comes to situations like these. Nevertheless, I look ahead to save myself work and immediately deal with a problem like the aforementioned, before it can fester into something more malevolent.
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The ! doesn't distribute. Almost any example will show this. For instance 4! - 3! = 24 - 6 =- 18, while (4 - 3) ! = 1! = 1.
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*******
- After look a bit ahead, I can see that my first instinct was true (to which I shall pat myself on the back as well as give myself a high-five) and continue on with the problem which is now read as:
(10!) / (4! * 6!)
- On a more succinct note, I should mention that I routinely will add parentheses (where they are usually unnecessary in this type of format) to both the numerator and denominator of fractions as a practice of keeping up with what I would input into a calculator to show that (10!) is the numerator and (4! * 6!) is the denominator.
- (10!) / (4! * 6!) should become something that looks like this:
(10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)
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([4 * 3 * 2 * 1] * [6 * 5 * 4 * 3 * 2 * 1])
- I take note that I can whack a big portion of the numerator away, corresponding to the equivalencies in the denominator below. (Specifically, [6 * 5 * 4 * 3 * 2 * 1])
- Now the fraction becomes something like:
(10 * 9 * 8 * 7
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- [4 * 3 * 2 * 1]
- From here, we get to attach some values to the other depending on which number goes best with which. For example, 10 and 2 would go better together than 10 and 4. (Yes, they’re both even numbers, but one would end up with something along the lines of 2.5 as a remainder)
- I divide 10/2, 9/3, 8/4, and 7/1.
- So now I have 5 * 3 * 2 * 7
- Which becomes 15 * 14
- And now we have the product and solution of the problem, 210.
confidence rating #$&*: 3
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Given Solution:: ** Starting with 10! / [ 4! * (10-4) ! ], we replace (10 - 4) by 6 to get 10! / ( 4! * 6! ).
Writing out the factorials our expression becomes
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 separatedly then divided out but it's easier and more instructive to divide out like terms.
Dividing ( 6 * 5 * 4 * 3 * 2 * 1) in the numerator by the same expression ( 6 * 5 * 4 * 3 * 2 * 1) in the denominator leaves us
10 * 9 * 8 * 7 / (4 * 3 * 2 * 1).
Every factor of the denominator divides into a number in the numerator without remainder:
Divide 4 in the denominator into 8 in the numerator,
divide 3 in the denominator into 9 in the numerator and
divide 2 in the denominator into 10 in the numerator and you end up with
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 divide 3628800 by 16480. But that process would lose accuracy and be ridiculously long for an expression like 100 ! / ( 30! * 70!). For this calculation the numbers involved would be hundreds of digits long. Much better to simply divide out like factors until the denominator goes away, as it always will with expressions of this form. (form n ! / (r ! * (n - r)! ). **
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Self-critique (if necessary): Pretty good. Feel free to laugh at any point during that problem; but pay more attention to my personal note here about the interesting way to divide out the numbers after eliminating 6!.
I certainly enjoyed that one.
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Self-critique Rating: ok
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Very good. Glad you avoided the malevolence.
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question 10.2.27 (previously 11.2.31) (10th edition #25) 3 switches in a row; fund count prin to find # of possible settings
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Your Solution:
3 switches with 2 settings:
2 * 2 * 2 = 8 possible settings
confidence rating #$&*: 3
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Given Solution:: **
There are two possible settings for the first switch, two for the second, two for the third.
The setting of one switch is independent of the setting of any other switch so the fundamental counting principle holds.
There are therefore 2 * 2 * 2 = 8 possible setting for the three switches.
COMMON ERROR: There are six possible settings and I used fundamental counting principle : first choice 3 ways, second choice 2 ways and third choice 1 way or 3 times 2 times 1 equals 6 ways.
INSTRUCTOR CRITIQUE: You're choosing states of the switches and there are only two states on each. No single choice has 3 possibilities. Also the setting of each switch is independent of the settings of the others, so the number of possibilities on one choice is independent of the number of possibilities on the other. Each of the 3 choices therefore has 2 possibilities. **
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Self-critique (if necessary): None.
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Self-critique Rating: ok
question 10.2.29 (previously 11.2.31) (10th edition #27) If no two adjacent switches are off why does the fundamental counting principle not apply?
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Your Solution:
- I don’t particularly follow what the question pertains to, but I accept the parameters of the FCP here.
- If any event in the problems given proves to be dependent on another variable within the problem, then the FCP does not apply.
- In other words, we can’t have the events within the material to interfere with the other variables the problem may have.
confidence rating #$&*: 2
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Given Solution:: ** The reason the principle doesn’t apply is that the Fund. Counting Principle requires that the events be independent. Here the state of one switch influences the state of its neighbors (neither neighbor can be the same as that switch). So the choices are not independent.
The Fund. Counting Principle requires that the events be independent. Since this is not the case the principle does not apply. **
STUDENT QUESTION
I thought for sure I had this right, but apparently I am confused. A switch is considered
independent when it is concerned with the placement of the other switches. It is dependent if it is not concerned. These
terms seem backward to me.
INSTRUCTOR RESPONSE
The given condition is that no two adjacent switches can be off.
So for example if the first switch is off, the second switch cannot also be off.
Or if the middle switch is off, neither of the other two can be off.
Thus in some circumstances the state of one switch has an influence on the state of the other switches, meaning that its state is dependent on the state of the other switch, and the Fundamental Counting Principle does not apply.
I believe that in this case, you did simply reverse the meanings of 'dependent' and 'independent'.
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Self-critique (if necessary): What is another circumstance that one may deduce whether or not the fundamental counting principle does or does not apply? I’m interested to see a contrasting example.
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Another direct analogy would be lining up a mixed group of people with the restriction that no more the two consecutive people can be of the same gender.
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Self-critique Rating: ok
question 10.2.40 (previously 11.2.36) How many odd 3-digit #'s from the set {3, 4, 5}?
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Your Solution:
- In my expanded 11th edition, this problem takes place on (11.2.42)
- I accept that the solution is thus
3 * 3 * 2 = 18
because:
- There are 3 numbers from which one could possible put in a box, also 3 numbers that one could put in the second box, but only 2 numbers (3,5) that are odd to be placed in the last.
- ***Why are there also 3 options to be put in the second box?***
confidence rating #$&*:
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Given Solution:: ** Using the box method, where in this case we draw three boxes and put one number in each:
The 1st box can be filled with any of the three so the first number of possibilities is 3
The 2nd box can also be filled with any of the three so the second number of possibilities is 3
The last digit must be odd, so there are only 2 choices for the third box.
By the Fundamental Counting Principle we therefore have 3*3*2=18 possible combinations.
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Self-critique (if necessary): Ok. Wasn’t aware any of the numbers could be used multiple times. Please elaborate either on your solution. See my bold print/underlined response under “Your Solution”
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Self-critique Rating:
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Note that bold and underline do not come through in a text form.
I assume you are referring to your question about the 3 options for the second box. It appears that you've understood the answer to that question.
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question 10.2.54 (previously 11.2.56) (10th edition 50) 10 guitars, 4 cases, 6 amps, 3 processors; # possible setups
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Your Solution:
- I’m going to work under the assumption that since there are four different variables here, that there will be 4 options to choose from.
- 10 * 4 * 6 * 3 = 40 * 18 = 720 possible setups
-
confidence rating #$&*:3
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Given Solution:: A setup consists of a guitar, a case, an amp and a processor.
There are 10 choices for the guitar, 4 choices for the case, 6 choices for the amp and 3 choices for the processor.
So by the Fundamental Counting Principle there are 10 * 4 * 6 * 3 = 720 possible setups. **
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Self-critique (if necessary): OK
You may add comments on any surprises or insights you experienced as a result of this assignment:
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Self-critique (if necessary): Speaking from the experience of someone who has usually hated fractions, I am pleased with the problem given with a new technique to eliminate numbers in a safe and effective manner.
I do, however, have a few comments regarding the questions.
1. With regards to the switch problem question 10.2.27 (previously 11.2.31) how are we to treat the basis of the problem? Are we to assume that there’s a light the switches designate is off or on? Or just worry about the switches will directly affect another, and thus dub them “dependent” which renders the FCP useless?
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The question was just about the sequence of the switches, not the effect of the sequence.
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2. Why does their independent nature cause the FCP to not apply? Why does the dependent nature of variables allow the FCP to apply to them?
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The FCP applies only to independent events. If there is any dependency, things get much more complicated.
Think of a path branching off into 3 paths, each of which branches off into 4 paths. There are clearly 3 * 4 = 12 possible routes.
However if there is some rule that restricts your choices, say a rule like you can't take one of the leftmost paths if you to the leftmost path the first time, then there is no simple multiplication that gives you the number of possible routes. You have to analyze the different cases.
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3. With regards to question 10.2.40 (previously 11.2.36) how are we to treat the problem when the situation, like in this question, seems to be vague? I’m comparing the problems to the ones you construct in your QA sections where you directly list a certain parameter that must be met.
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I don't see that question here and I don't carry a copy of the text. I'll be glad to answer but I need a copy of the question.
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- I feel that if the problem specifically stated “How many odd numbers can one form from the set {3, 4, 5} where there are no constraints to repetition or order?” I would immediately say to myself “Hey, I’ve got a free-form question here where I can say the first number can be any of the 3 and the second number can be any of the 3 but man, that last number has to be either a 3 or a 5 because if it was a 4 that would lead to the total number of combinations, not just the odd numbers”
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The FCP applies in this case.
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4. I’ve always been told it’s dangerous to assume anything in mathematics without either going by a law, theorem, or special circumstance which dictates, word for word, a particular rule or understanding (such as how one would treat x(3-4) = 3x - 4x, by the law of distribution or how one assumes that the denominator of a whole number is always 1)
- Is it dangerous for me to assume that, if I was given a problem like
10 clarinets, 9 flutes, 3 saxophones, 5 piccolos, and 4 oboes are in a woodwind section of a symphony. 31 players can all play any instrument interchangeably. How many combinations can be made between each player and an instrument?
- Where I would assume that they could appear in any order (as each player can play any of the instruments) that I would simply multiply 10 * 9 * 3 * 5 *4?
- I would note, though, that I’m thinking this would be a case where they are unordered (and thus I would divide that product by 5! ( 5 * 4 * 3 * 2 * 1) to arrive at the solution of 45 different combinations)
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This is a little more complicated. Questions of this type are coming up.
There would be C(31, 10) ways to assign 10 players to the clarinets, then C(21, 9) ways to assign 9 of the remaining player to flutes, then C(12, 3) ways to assign the remaining twelve players to the saxes, etc.. You would the apply the FCP again and multiply all these results.
You would get a large number.
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Self-critique Rating: Ok, see comments above
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Lots of good thinking here. Also imagination. Good things.
Check my notes.
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