permutation

Question

at times it becomes confusing...
The total numbers of ways in which 7 balls can be distributed amongst 9 persons = 9^7 ways ie each ball can go to any of 9 guys..... but if we take each man can take any of seven balls ans wud be 7^9

abhi2109 · Answer

One person can take none,one or more than one ball however one ball cannot be distributed to two persons, choice is with balls.....ans is 9^7.

meatdumpling · Answer

what about the formula? I dont' recall exponents

9 NCR 7, but since this is permutations, not combinations (the order matters and there's more), then we don't divide by 7!

So 9! / 2! = 3*4*5*6*7*8*9 = Some BIG number

But if the balls are all identical then it's not permutations. It's combinations. In that case e have (9! / 2!) / 7!
= 8*9 / 2 = 36

Is that right?

brandy96 · Answer

can anybody explain this in little detail?

jko · Answer

The solution here depends on whether a single person can have more than one ball.

If any given person can have more than one ball, I have no idea how to solve that. The combinations are almost limitless.

If each person can have only one ball max, meatdumpling's answer is correct.

victorxman182 · Answer

If each person can have 0, 1 or more balls, the choice is on the balls i.e. Ball #1 has 9 choices (each guy), Ball #2 has 9 choices too (the same guys), this holds true for the remaining balls. 
Therefore the total numbers of ways in which 7 balls can be distributed amongst 9 persons is 9*9*9*9*9*9*9 = 9^7

victorxman182 · Answer

This is not true because a ball can't belong to two or more persons, only one. I think the statement needs a better explanation of what "distribute" means. It can mean that each person can take only one ball, or that each person can take more than one balls. It can also mean that balls can be left without a person assigned or that each ball must have a person assigned.

meatdumpling · Answer

Ummm, I still think my 36 is a better answer than these exponents you're throwing out

Gladius77 · Answer

I have a basic permutation question where I need some help in getting to the answer fairly quickly. 
Question: If P(n,3) = 210 , find value of n?

So here is what I am doing: n!/(n-3)! = 210
n x (n-1) x (n-2) = 210

I know the answer should be 7 x 6 x 5, but what's the easiest way to split 210 and get to value of n? More importantly how do you know it's 7,6&5? Any help is appreciated.

GMATPill · Answer

There is confusion here as the question is being interpreted in ways not mentioned in the question. Be careful.

You use 7^9 if:  each person can pick one of those 7 ball types and those types can be repeated. An analogy would be a slot machine with 7 items and 9 slots. A red ball showing in slot #1 may or may not show up in slot #5. Here you would use 7^9 or 40+million

You use 9C7 (9 choose 7 for COMBINATIONS) if:  each person can pick one of those 7 balls --each person can only have maximum one ball. The formula is 9! / ((9-7)! 7!)= 36

You use 9P7 (9 pick 7 for PERMUTATIONS) if:  each person can pick one of those 7 balls and each of those 7 balls were a different color. In a combination, the following is considered one combination: the first 7 guys have a ball but the last two don't. But In a PERMUTATION, this single order could translate into a bunch of permutations with the different colored balls. The first 7 guys still have the balls and the last two don't. Same scenario. Except now these first 7 guys can form a bunch of different color combinations in different orders. #1 can have white ball, #2 can have red ball, etc; and then #1 can have red ball and #2 can have white ball, etc. In this case, the formula is: 9! / (9-7)! = 181,440 (a very large number)

So you can see there's 3 ways to look at this question. Be careful on variations/combinations/permutations.

GMATPill · Answer

So basically the "3" tells you how many terms you need to multiply because the rest of the terms cancel out.
For example, if you picked 8, then you would test 8*7*6.
If you picked 7, then you would test 7*6*5.

The reason you do this is because the 4! from (7-3)! on the denominator cancels out with the 4! inside the 7! on the numerator. SO you're left with the numbers between 7 and 4--which are 7*6*5.

So basically you do guess and check. Try 5*4*3---if it's too low, then try 8*7*6. If that's too low, then try 7*6*5 and you'll get 210.

Gladius77 · Answer

Thanks for your reply. I see what you mean about guessing. So let's say we try another similar question -

Question: If P (m+n,4) = 3024 and P(m-n,4) =120, find m & n? How would you attack this problem? Is there any non-guessing way of deriving the solution?

sdas · Answer

Hi can any1 recomment combinatorics & probability concept books for GMAT apart from MGMAT....

permutation

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