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Many of you may have seen my other blog "You guessed it, Math Homework." Well, you guessed it, MATH HOMEWORK!
Since this chapter is self-evaluation (Meaning we learn it ourselves), and I don't understand it, I've come to TL for answers.  (Doesn't everyone? XD)
So I'm supposed to do my homework using nCr and nPr (Permutations and Combinations) as needed. I'm like 99% sure that these problems require permutation, except (like last time) I don't know how to set up the equation. T_T
Janice has 8 DVD cases on a shelf, each case is one season of CSI. Her brother accidently knocks over all the cases and puts them back on the shelf randomly.
#1. P(Seasons 1-4 in the correct positions) So 1234XXXX
I got 1/70 for this one. I seriously doubt it's correct though.
#2. P(Seasons 1 and 8 in the correct positions) 1XXXXXX8
Currently stuck on this one.
 
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#1. P(Seasons 1-4 in the correct positions) So 1234XXXX
= number of solutions/total number of positions
can you figure out the total number of positions?
can you figure out the total number of solutions?
it's really easy, just think about it.
#2. P(Seasons 1 and 8 in the correct positions) 1XXXXXX8
Currently stuck on this one.
same thing.
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yeah this type of stats is common sense. you can find the solutions pretty easily without using choose or permute. just do exactly what ydg said: find # of solutions, # of total positions, and work it out
if you're still having trouble, post back your thought process and we can see where (if anywhere) you go wrong
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For #2 I tried P(8,2)/P(8,6)+P(8,2)
P(8,2) because 8 total DVD's and 2 are in the correct positions and then P(8,6) equals the number that are in the wrong order (failure rate)
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I don't understand how to find # of solutions though.
Like I know the total # of positions is 8! = 40320 which will be the denominator...but then the numerator D=
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On February 23 2009 09:24 OmgIRok wrote:
I don't understand how to find # of solutions though.
Like I know the total # of positions is 8! = 40320 which will be the denominator...but then the numerator D=
well ok
what constitues a solution in part 1? 1234XXXX
how many ways can you arrange ur dvd's such that this is the case
so you have
1234 5678
1234 5687
etc.
keep thinking you almost got it
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On February 23 2009 09:25 JeeJee wrote:Show nested quote +On February 23 2009 09:24 OmgIRok wrote:
I don't understand how to find # of solutions though.
Like I know the total # of positions is 8! = 40320 which will be the denominator...but then the numerator D= well ok what constitues a solution in part 1? 1234XXXX how many ways can you arrange ur dvd's such that this is the case so you have 1234 5678 1234 5687 etc. keep thinking you almost got it
Hmmm would it be p(8,4) = 1680? and then divide that by 8! so 1/24?
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That's not quite right. You want to count the number of arrangements 1234xxxx. How many ways can you put 5,6,7,8 into xxxx?
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um 4! = 4*3*2*1 = 24 ways?
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On February 23 2009 09:32 OmgIRok wrote:Show nested quote +On February 23 2009 09:25 JeeJee wrote:On February 23 2009 09:24 OmgIRok wrote:
I don't understand how to find # of solutions though.
Like I know the total # of positions is 8! = 40320 which will be the denominator...but then the numerator D= well ok what constitues a solution in part 1? 1234XXXX how many ways can you arrange ur dvd's such that this is the case so you have 1234 5678 1234 5687 etc. keep thinking you almost got it Hmmm would it be p(8,4) = 1680? and then divide that by 8! so 1/24?
yes.... since they first 4 are already in position.... u just need to worry about the last 4.... soo 4P4 or 4!
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so the success rate is 8! - 4! and the denominator is 8!
edit: um I know how to get the total solution, but probability is succesful possibilities/total possibilities, and so if i find out how many ways 5678 can be arranged (which is 4!) thats the success rate? I'm pretty sure i have to multiply that to something...
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huh?? you question is the number of ways u can arrange the last 4 cases right?
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On February 23 2009 09:41 Monoxide wrote:
huh?? you question is the number of ways u can arrange the last 4 cases right?
the question asks whats the probability that the 8 cases are put back so that 1234 are in the correct positions, cases 5678 dont matter
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On February 23 2009 09:39 OmgIRok wrote:
so the success rate is 8! - 4! and the denominator is 8!
edit: um I know how to get the total solution, but probability is succesful possibilities/total possibilities, and so if i find out how many ways 5678 can be arranged (which is 4!) thats the success rate? I'm pretty sure i have to multiply that to something...
You've pretty much got it with "successful possibilities/total possibilities"
The ways 5678 can be arranged is your successful possibilities.
Total possibilities is the number of ways you can arrange 12345678.
The second question is very similar. You can use nCr or nPr if you find it easier that way, but all you really need is there are x! permutations of x objects. (But you should understand what nCr nPr actually do, you'll realise you might not want/need it)
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On February 23 2009 09:51 Nytefish wrote:Show nested quote +On February 23 2009 09:39 OmgIRok wrote:
so the success rate is 8! - 4! and the denominator is 8!
edit: um I know how to get the total solution, but probability is succesful possibilities/total possibilities, and so if i find out how many ways 5678 can be arranged (which is 4!) thats the success rate? I'm pretty sure i have to multiply that to something... You've pretty much got it with "successful possibilities/total possibilities" The ways 5678 can be arranged is your successful possibilities. Total possibilities is the number of ways you can arrange 12345678. The second question is very similar. You can use nCr or nPr if you find it easier that way, but all you really need is there are x! permutations of x objects.
But for the 2nd question I'm pretty sure combinations wont help me, and i tried P(8,2)/8! which gave me 1/720
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second question? 6! / 8! ??
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On February 23 2009 09:58 OmgIRok wrote:Show nested quote +On February 23 2009 09:51 Nytefish wrote:On February 23 2009 09:39 OmgIRok wrote:
so the success rate is 8! - 4! and the denominator is 8!
edit: um I know how to get the total solution, but probability is succesful possibilities/total possibilities, and so if i find out how many ways 5678 can be arranged (which is 4!) thats the success rate? I'm pretty sure i have to multiply that to something... You've pretty much got it with "successful possibilities/total possibilities" The ways 5678 can be arranged is your successful possibilities. Total possibilities is the number of ways you can arrange 12345678. The second question is very similar. You can use nCr or nPr if you find it easier that way, but all you really need is there are x! permutations of x objects. But for the 2nd question I'm pretty sure combinations wont help me, and i tried P(8,2)/8! which gave me 1/720
P(8,2) means the number of ways of picking 2 objects from 8, where order does matter, that doesn't really help.
Okay I was being vague, but what you're really doing is fixing 1 and 8. And permuting the 6 in between. So you're really doing P(6,6) = 6!.
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wait yeah order does matter because 1 and 8 have to be in the correct positions
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On February 23 2009 10:07 Nytefish wrote:Show nested quote +On February 23 2009 09:58 OmgIRok wrote:On February 23 2009 09:51 Nytefish wrote:On February 23 2009 09:39 OmgIRok wrote:
so the success rate is 8! - 4! and the denominator is 8!
edit: um I know how to get the total solution, but probability is succesful possibilities/total possibilities, and so if i find out how many ways 5678 can be arranged (which is 4!) thats the success rate? I'm pretty sure i have to multiply that to something... You've pretty much got it with "successful possibilities/total possibilities" The ways 5678 can be arranged is your successful possibilities. Total possibilities is the number of ways you can arrange 12345678. The second question is very similar. You can use nCr or nPr if you find it easier that way, but all you really need is there are x! permutations of x objects. But for the 2nd question I'm pretty sure combinations wont help me, and i tried P(8,2)/8! which gave me 1/720 P(8,2) means the number of ways of picking 2 objects from 8, where order does matter, that doesn't really help. Okay I was being vague, but what you're really doing is fixing 1 and 8. And permuting the 6 in between. So you're really doing P(6,6) = 6!.
yes... soo 6! / 8!
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EPT
