nPr and nCr

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atlanta213

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For the probability question in QR, How do we know when we need to use nPr or nCr?


Thanks

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For the probability question in QR, How do we know when we need to use nPr or nCr?


Thanks

It has to do with whether or not order matters. For example, if a teacher is wondering how many different groups she can divide her class into, order wouldnt matter. The group with Bob, John, Mike would be the same even if they were listed as John, Mike, Bob. This is a combination.

On the other hand, if the teacher wants to know how many different groups of 1st, 2nd, and 3rd place finishers her class can have in a spelling bee, order would matter. For example if Bob - 1st, John - 2nd, Mike - 3rd thats a DIFFERENT group than John- 1st, Mike - 2nd, Bob - 3rd. This is a permutation.

See how ORDER MATTERS in the second example but not the first?? If not, post some problems and I can help you out. It is 8:30am so forgive me if my examples arent awesome haha.
 
It has to do with whether or not order matters. For example, if a teacher is wondering how many different groups she can divide her class into, order wouldnt matter. The group with Bob, John, Mike would be the same even if they were listed as John, Mike, Bob. This is a combination.

On the other hand, if the teacher wants to know how many different groups of 1st, 2nd, and 3rd place finishers her class can have in a spelling bee, order would matter. For example if Bob - 1st, John - 2nd, Mike - 3rd thats a DIFFERENT group than John- 1st, Mike - 2nd, Bob - 3rd. This is a permutation.

See how ORDER MATTERS in the second example but not the first?? If not, post some problems and I can help you out. It is 8:30am so forgive me if my examples arent awesome haha.

Nicely explained, I always have a difficult time trying to explain this. You did a pretty darn good job.
 
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greenwave, i totally understand your explanation.

can you apply your reasoning to the actual question? and explain again?

Thansk
 
nPr = P!/r!(P-r)! and nCr= p!

Is this correct???? thanks!

also circular permutations are good to know, its just (n-1)!
 
nPr = P!/r!(P-r)! and nCr= p!

Is this correct???? thanks!

Not quite man.

nPr = n! / (n-r)!
nCr = n! / r! (n-r)!

The way to remember which formula is combo vs permutation that works for me is - if order matters (permutations) there will be more possible groups (ill illustrate this in a sec). So if there are more possibilities then you will divide by less stuff.
 
greenwave, i totally understand your explanation.

can you apply your reasoning to the actual question? and explain again?

Thansk

Ok sure, ill use my to previous examples.

1) For example, if a teacher is wondering how many different groups of 3 she can divide her class of 10 students into, order wouldnt matter. The group with Bob, John, Mike would be the same even if they were listed as John, Mike, Bob. This is a combination.

Once you realize this is a combination, you need to identify "r and n." (in the formula there are "n" number of things picked "r" at a time). So there are 10 students (this is the number of things) picked 3 at a time (form groups of 3). Which means n=10 and r=3. So:

nCr = n! / r!(n-r)!
10 C 3 = 10! / 3!(10 - 3)!
10 C 3 = 10! / 3! (7)!
10 C 3 = 10 X 9 X 8 X 7! / 3! (7!) now you can cancel the 7! from top and bottom
10 C 3 = 10 X 9 X 8 / 3 X 2 = 120 different groups of 3

2) On the other hand, if the teacher wants to know how many different groups of 1st, 2nd, and 3rd place finishers her class of 10 students can have in a spelling bee, order would matter. For example if Bob - 1st, John - 2nd, Mike - 3rd thats a DIFFERENT group than John- 1st, Mike - 2nd, Bob - 3rd. This is a permutation.

Now identify n and r again. Since there are 10 students picked for groups of 3 ( 1st, 2nd, and 3rd), n and r are the same as above.

n P r = n! / (n-r)!
10 P 3 = 10! / (10 - 3)!
10 P 3 = 10! / 7!
10 P 3 = 10 X 9 X 8 X 7! / 7! you can cancel the 7! from top and bottom again
10 P 3 = 10 X 9 X 8 = 720

** big difference eh?? hope that helped :)
 
if order matters, there more possibilities?
its so counterintuitive.

think about my examples. if order WITHIN the group doesnt matter then John + Mike + Bob is the same group as Mike + Bob + John which is also the same group as Mike + John + Bob.

but when order within the group matters (permuatations) there are so many more possibilities. think of my second example 1st place - John, 2nd place - Mike, 3rd place - Bob is a DIFFERENT group than 1st place Mike, 2nd place - Bob, 3rd place - John which is a DIFFERENT group than 1st place - Mike, 2nd place - John, 3rd place - Bob.

you see? when order within the group matters there are so many more options
 
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for circular permutations let me give an example:

What is the arrangement of 5 individuals around a circular table?

(5-1)! = 4x3x2x1=24

you just subtract 1 and do factorial pretty simple...
 
for circular permutations let me give an example:

What is the arrangement of 5 individuals around a circular table?

(5-1)! = 4x3x2x1=24

you just subtract 1 and do factorial pretty simple...


does this example take into account the order that people are sitting around the table - if not, then its a combination right?

also, ok so in regards to circular permutation - what if the question is something like - how many ways can you order 5 pple around a table in which order matters - do u use the same formula or something different?
 
"Circular permutation" doesn't have a formula because there are problems where you can't rely on a single formula. If you straight up have n people sitting at a round table then yes it boils down to (n-1)! but if the problem is for example the following:

"You have 7 people and want to seat 5 of them at a round table. How many ways can this be done?"

... then you cannot do the formula. You have to reason it out. In this case you'd have to choose the people sitting down using a permutation (7 P 5) = 2520. Then since there are 5 seats available, you divide by 5 since the number 2520 accounts for the instances where everyone simply shifts over an identical # of seats but keeps the order around the table. The person at seat 1 could have been at seats 2 through 5 with the same order going around clockwise. Thus only 1 out of every 5 arrangements are UNIQUE with regard to order. So you really have 2520/5 = 504 unique seating arrangements.

Point is, you need to reason it out in your head.



Flycd05 - the table problems are permutations always if it wants you to select where people sit in relation to each other. If it asks you to find the # of ways to select people to be at a table, then that's a combination because it doesn't care about where they sit. You need to ask yourself when deciding between combo and perm, 'does it matter HOW I choose the people in each group (perm), or does it only matter that each group is unique from the rest (combo)?'
 
A house has 4 rooms, and a painter has 9 colors. If the rooms are A, B, C, and D, how many different ways can he paint the rooms?

I having a hard time figuring out whether this is a nPr or nCr..
The book saids nPr, but cant figure out why? Is it b/c they lettered in a specific way like "A,B,C, and D?"
 
Well I won't give you the answer because I gave that explanation above so I want you to think. Remember that permutation means order is important and combination means that order is not important.

You need to ask yourself a question. When choosing the 4 colors, does it matter if the four colors are associated with specific rooms?
 
yea..the answer is 9!/(9-4)! = 3024 which is the Permutation formula meaning order does matter.

I guess when it says "how many different ways", its referring to the order..

If this "was" a Combination problem, how would it be worded? This would help me out alot. THANK YOU! :D:D
 
A house has 4 rooms, and a painter has 9 colors. If the rooms are A, B, C, and D, how many different ways can he paint the rooms?

I having a hard time figuring out whether this is a nPr or nCr..
The book saids nPr, but cant figure out why? Is it b/c they lettered in a specific way like "A,B,C, and D?"

I think so because that's like saying there are 4 prizes and 9 candidates. If prizes are 1st, 2nd, 3rd, and 4th places, how many diff ways can these ppl be awarded...something like that :)
 
That's because
A=1, B=1, C=1, D=2 and
A=2, B=1, C=1, D=1 are different

thus order matters

nPr = order is imPortant
nCr = order don't Care
 
this is destroyer question 87.

Al, Bob, Kathy, Ed, Scot and Frank are running for office. They need a committee to consist of a president, a vive president, a secretary, and a treasurer. How many different committees can consist of these office?

I thought that order does not matter for this problem so, I did 6!/[4!(6-4)!]

BUT it turns out that order matter..

I don't get it

please help!!

Thanks
 
this is destroyer question 87.

Al, Bob, Kathy, Ed, Scot and Frank are running for office. They need a committee to consist of a president, a vive president, a secretary, and a treasurer. How many different committees can consist of these office?

I thought that order does not matter for this problem so, I did 6!/[4!(6-4)!]

BUT it turns out that order matter..

I don't get it

please help!!

Thanks
Man think it through. Would things be different if Cheney was Prez and Bush was VP? I do think so!
 
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Man think it through. Would things be different if Cheney was Prez and Bush was VP? I do think so!

omg.... yea you are right...

I am just out of it today..

my DAT is in like less than 2 weeks.. and i've been just studying snd studying...

any ways Thanks!!!
 
How about this question?
It is from ADA sample DAT QR...

Jill has 6 different books. In how many ways, can she choose two different books?

The answer is 30. There is no solution but I am assuming 6P2 is used here.

But in my approach even though there are two different books, the order shouldn't matter.For example, there are 6 books - A, B, C, D, E, F. In choosing A first and B or B first and A, the order should not matter.. right?
So doesn't it have to be 6C2 rather than 6P2??


This question seems so simple but I don't know why I got confused....
 
well order should matter since they are different books, so choosing book A first and B second would be a different way than choosing B then A, so its 6!/(6!-2!) = 30
 
well order should matter since they are different books, so choosing book A first and B second would be a different way than choosing B then A, so its 6!/(6!-2!) = 30


Really? but after choosing two books, you will have just two books A and B on your hands..

Is this the similar situation of seleting a president and a vice president from six people? It makes sense in this case but still confused about the case w/ books....:(
 
She can choose book A and then reach out and choose book B... or the reverse.

How many ways can she select two books to read? That would be nCr.
How many ways can she select two books to read, one per week? I'd go with nPr there.
 
These explanations were really helpful....im not currently studying the math right now, but reading this just made probability make more sense to me. Thanks guys.
 
I was confused with these type of problems but now reading the responses above really helped. thanks guys! and good luck on DAT :luck:
 
Old topic but a good one....

Can someone explain these 2 examples?

1) Find the number of ways that 6 objects A,B,C,D,E,F can be taken 3 at a time.

- answer: P= nPr = 6!/3!
- Why use nPr instead of nCr?

2) A weatherman says that theres a 70% chance of rain for 3 days. What is the probability of raining 2 out of 3 days?

- answer: P=(nCr)(p^n)(1-p)^r = (3!/2!1!)(7/10)^2(3/10)^1 = 441/1000
 
Old topic but a good one....

Can someone explain these 2 examples?

1) Find the number of ways that 6 objects A,B,C,D,E,F can be taken 3 at a time.

- answer: P= nPr = 6!/3!
- Why use nPr instead of nCr?

2) A weatherman says that theres a 70% chance of rain for 3 days. What is the probability of raining 2 out of 3 days?

- answer: P=(nCr)(p^n)(1-p)^r = (3!/2!1!)(7/10)^2(3/10)^1 = 441/1000

P=(nCr)(p^n)(1-p)^r

nCr only tells you how many possible combinations of ways you can get rain over the 3 days (0 days of rain + 1 day of rain + 2 days of rain + 3 days of rain). When you're asked a question where you want to filter out only a certain type of combination you have to multiply in two other factors: the probability of the favored even raised to the power = # times you want it to occur, and the probability of the unfavored event raised to the power you want it to occur.

i.e. if the question were 60%chance of rain, what's the probability of 5 days of rain out of 7?

(nCr)=7!/5!/2!= all combinations of rain/no rain for 7 days
p^n= (6/10)^5= probability of rain, and you want it to happen for 5 days
(1-p)^r= (4/10)^2= probability of no rain, and you want it to happen twice

*edit* oh, and if order matters, you'd use nPr, like if the wording was "how many ways can it rain 5 days out of the week?"
 
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from Destroyer, in ball selection problem, it says, the balls are drawn out "without replacement". What does this mean?
 
Man think it through. Would things be different if Cheney was Prez and Bush was VP? I do think so!

So it does matter what room is what color (room one is green, 2 blue different than one blue two green), which I agree with. But it doesn't matter who's president, who's VP? (I don't know about the politics of it, but Yes I think it's def a different situation if Cheney is pres and bush VP vs. the opposite, at least in math terms.
 
P=(nCr)(p^n)(1-p)^r

nCr only tells you how many possible combinations of ways you can get rain over the 3 days (0 days of rain + 1 day of rain + 2 days of rain + 3 days of rain). When you're asked a question where you want to filter out only a certain type of combination you have to multiply in two other factors: the probability of the favored even raised to the power = # times you want it to occur, and the probability of the unfavored event raised to the power you want it to occur.

i.e. if the question were 60%chance of rain, what's the probability of 5 days of rain out of 7?

(nCr)=7!/5!/2!= all combinations of rain/no rain for 7 days
p^n= (6/10)^5= probability of rain, and you want it to happen for 5 days
(1-p)^r= (4/10)^2= probability of no rain, and you want it to happen twice

*edit* oh, and if order matters, you'd use nPr, like if the wording was "how many ways can it rain 5 days out of the week?"


The way I answer that question is this. Chance of raining each day is 0.7 and not raining 0.3. Thus for only 2 out of 3 days to rain, the probability is:
(first 2 days rain or first and last day rain or last two days rain):
0.7*0.7*0.3 + 0.7*0.3*0.7 + 0.3*0.7*0.7 = 441/1000
 
So it does matter what room is what color (room one is green, 2 blue different than one blue two green), which I agree with. But it doesn't matter who's president, who's VP? (I don't know about the politics of it, but Yes I think it's def a different situation if Cheney is pres and bush VP vs. the opposite, at least in math terms.
Yeah, I said I DO think so.
 
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