permutation

can anybody explain this in little detail?

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.

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

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.

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

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.

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.

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.

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?

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