Math problem

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WasabiPoptart

WasabiPoptart

Last night I took a comprehensive quiz for my math course. I got a 95! WOO! But the one problem I got wrong is really bugging me:
There are 4 senators on a committee: Amy, Brian, Carol, and Dennis. Any or none of them may be chosen for other subcommittees. How many possible subcommittees are there?

I thought the solution was n^n (n to the nth power) which gave me (4^4) 256. But that isn't right. I can't review the correct answers until after the final is over. However, this is driving me crazy. Did I have the right idea or was I completely off?
 
Dei

Dei

Uh, if there can be subcommittees with none of them in it, that kind of makes the number infinite... unless I am reading that wording way off.
 
Allen who is Quiet

Allen who is Quiet

I think you had the right idea, but there's the possibility none of them are on the subcommittee, one of them is, two of them are, three of them are, or all of them are. So the number of possible ones should be:

(number of possible subcommittees containing none) + (containing one) + (containing two) + (containing three) + (containing all of them)
 
tegid

tegid

I just did the combinations on a piece of paper and got 16 as the result. Now, that's not useful without the math behind it, right? So I looked up on the internet the various formulas to help me explain and here it is:

The formula you used, n^n, is for permutations with repetition, i.e. the order matters and you can have the same senator any number of times, and you are always choosing exactly 4 of them. Maybe there's some other case the formula is good for...

In this case, what you need to use is the formula for combinations (the order doesn't matter) without repetition, which is n!/(r!(n-r)!), where n is the number of elements you choose from (in this case 4) and r the number of elements you pick (in this case it ranges from 0 to 4). So you'd get:
# of combinations of 0 senators + # of combinations of 1 senator + #of combinations of 2 senators + ... 3 senators + ... 4 senators =
4!/(0! 4!) + 4!/(1! 3!) + 4!/(2! 2!) + 4!/(3! 1!) + 4!(4! 0!) =
1 + 4 + 6 + 4 +1 = 16

(n!= n * n-1 * n-2 * ... * 2 * 1, and 0! = 1)

EDIT:

The trick to laying out all the combinations by hand is fixing the elements in order: you fix the first one and cycle through the rest:
all the possibilities with A
AB,AC,AD
now all the possibilites with B. Since we already accounted for A, we leave it out.
BC,BD
now all the possibilities with C. We leave out A and B:
CD
And that's it.

It may sound ridiculous but I'm not good with combinatorics and having some sort of system helps in more complicated cases.
 
Ravenpoe

Ravenpoe

Allen who is Quiet said:
I think you had the right idea, but there's the possibility none of them are on the subcommittee, one of them is, two of them are, three of them are, or all of them are. So the number of possible ones should be:

(number of possible subcommittees containing none) + (containing one) + (containing two) + (containing three) + (containing all of them)
Click to expand...

For the more visual:

  1. Amy, Brian, Carol, and Dennis
  2. Amy, Brian, Carol
  3. Amy, Brian, Dennis
  4. Amy, Carol, Dennis
  5. Brian, Carol, Dennis
  6. Amy, Brian
  7. Amy, Carol
  8. Amy, Dennis
  9. Brian, Carol
  10. Brian, Dennis
  11. Carol, Dennis
  12. Amy
  13. Brian
  14. Carol
  15. Dennis
  16. None of them
(1)+(4)+(6)+(4)+(1)
 
WasabiPoptart

WasabiPoptart

Thanks everyone! I was going to write it all out like Poe posted (visual is definitely the way to go for me on this) and decided not to since I left this problem for last and time was ticking away. I should have gone with it.
 
PatrThom

PatrThom

The binary method is the most visual, but you can use Tegid's method to calculate your own lottery odds.

--Patrick
 
MindDetective

MindDetective

Something seems off about that. Let's try swapping in some different values...

.33333...=x
9.33333...=10x
9.33333... - .33333...=9 and 10x - x=9x
9 = 9x
1 = x
Therefore .33333...=1[DOUBLEPOST=1355337448][/DOUBLEPOST]

Ah, you just reassert the assumption in step two is all.
 
MindDetective

MindDetective

Dave said:
That would be 3.33333...=10x

3.33333...-.33333...=3 and 10x-x=9x

3=9x

1/3 = x
Click to expand...
I was intentionally trying to break it, which is why I only changed the decimals. I wanted to create an inconsistency to see where things were off so that x = both 1 and 0.33333...
 
Chad Sexington

Chad Sexington

The 1/3 thing does illustrate the point perfectly, actually.

.3333... = 1/3 and 1/3 + 1/3 + 1/3 = 3/3 or 1. (while adding the decimals gives you .99999999...)
 
MindDetective

MindDetective

Chad Sexington said:
The 1/3 thing does illustrate the point perfectly, actually.

.3333... = 1/3 and 1/3 + 1/3 + 1/3 = 3/3 or 1. (while adding the decimals gives you .99999999...)
Click to expand...
Well, that was my point. It just is another version of the same problem. I wanted to see where the equality really occurred, so I broke it on purpose. It turns out all you are doing in that progression is reasserting that 1 = 0.99999... in step 2. Thus it doesn't actually demonstrate equality, it just appears to.
 
Chad Sexington

Chad Sexington

MindDetective said:
Well, that was my point. It just is another version of the same problem. I wanted to see where the equality really occurred, so I broke it on purpose. It turns out all you are doing in that progression is reasserting that 1 = 0.99999... in step 2. Thus it doesn't actually demonstrate equality, it just appears to.
Click to expand...
I'm not sure I follow? I set x=.99... and in step two I multiply both sides by 10. Solve for x and you get 1. I am not sure how this doesn't demonstrate equality?
 
MindDetective

MindDetective

Chad Sexington said:
I'm not sure I follow? I set x=.99... and in step two I multiply both sides by 10. Solve for x and you get 1. I am not sure how this doesn't demonstrate equality?
Click to expand...
Because step two is where you are essentially asserting that 9.99999... = 10. It isn't apparent because of the number of decimals you are dealing with. But when you break it on purpose like I did earlier:

.33333...=x
9.33333...=10x
9.33333... - .33333...=9 and 10x - x=9x
9 = 9x
1 = x
Therefore .33333...=1

You can see that in step two 9.33333...=10x is essentially false, given our x initial x value. So at this step of the proof I have to reassert that these are in fact equal (thus ignoring my initial given value) and continue on from there. You are doing the same thing in your version as well, but it is cleverly hidden in the millionth decimal place somewhere.
 
Chad Sexington

Chad Sexington

No, I'm only saying that if you multiply .9999... by ten, then the decimal moves one place over (if you multiply 10.0 by 10, you get 100). I'm not stating that 9.999...=10 I'm stating 9.999... is equal to 10 times x which is .9999...

It does so happen that because .999... is equal to 1 that 9.999... is equal to 10, but the math isn't flawed.

Your setup makes no sense. In step two you don't multiply both sides by 10. You add 9 to the left and multiply the right by 10. That's not equivalent.
 
Chad Sexington

Chad Sexington

We are in agreement that .999... does in fact equal 1, right? You're just disputing my methodology?

When you do your method the way I just posted, to get 1/3 = x that also proves .9999... = 1. I also disagree that my previous method is wrong, however. I am not adding nine. I'm multiplying both sides by the same value.
 
MindDetective

MindDetective

Chad Sexington said:
We are in agreement that .999... does in fact equal 1, right? You're just disputing my methodology?
Click to expand...


No, I'm not convinced of that.

Chad Sexington said:
When you do your method the way I just posted, to get 1/3 = x that also proves .9999... = 1. I also disagree that my previous method is wrong, however. I am not adding nine. I'm multiplying both sides by the same value.
Click to expand...
Not really. You are also adding 9 to one side and multiplying the other by 10.
 
Chad Sexington

Chad Sexington

MindDetective said:
Did you read the discussion bit on Wikipedia? That's what we're fundamentally going round and round about here.
Click to expand...
I agree that it isn't a good explanation of why .999...=1 but the discussion section does state that it does demonstrate it. I am surprised you don't agree with the idea that .999...=1(in some number systems you can have infinitely small repeating decimals but not in the real number system).

I will try to come up with better sources/explanation if I can but I'm at work and should probably, you know, do some of that.
 
MindDetective

MindDetective

Well, it definitely approximates it. In fact, it is the best approximate of it after the number 1 itself (since we don't have a way to easily represent 1.000...1). We simply don't differentiate at such a fine level, so we round up.[DOUBLEPOST=1355341132][/DOUBLEPOST]
stienman said:
Does 0.9999... * 10 == 0.9999... + 9?
Click to expand...
That's the fundamental question, isn't it?
 
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