Permutation and Combination Conceptual Question

Question

5 balls are to be put in 3 boxes. In how many ways can this be done if:-
a)Balls are identical and boxes are different
b)balls are different and boxes are identical
c)both boxes and balls are different
d)both boxes and balls are identical
in which of the above cases the formulae (n+r-1)C(r-1) be applied?

Please explain WITH EXPLANATIONS. I am not very good at this topic. Thanks
I had one of the question in the CAT books. But this raised doubt about similar four cases. Kindly help me as i m way too confused.

KarishmaB · Answer

I have discussed all these cases taking the example of '5 fruits distributed among 4 children' in the following two posts: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/12 ... 93-part-1/ https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/12 ... s-part-ii/ You use (n+r-1)C(r-1) in case of n identical items distributed in r different groups - no restrictions You use (n-1)C(r-1) in case of n identical items distributed in r different groups - no group can be empty Its not advisable to deal with formulae. Try to understand the fundamentals as explained in the posts.

gmatkiller2015 · Answer

Thanks a lot karishma. I am planning to take GMAT around August-September, 2013. I am looking for admission in Indian B schools only for now as the cost of mba in usa is out of the roof. I am currently located in Pune, India. Please tell me about different veritas prep programs that are available (both online/classroom).

KarishmaB · Answer

Check out this link: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/ It will tell you a bit about our courses and will also provide you links to check out one of our books, one of our lessons and one of our practice tests for free. For the courses we offer, check this link: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/gmat-course-options/ In Pune, you have three options available: Online full course, Private tutoring (through me) and On Demand. The link gives you details of all three study options. Feel free to get back with any questions you might have, either here or on PM.

goyal1109 · Answer

These are typical questions which I like to convert to a simple arrangement / permutations problem. Assume that there are 2 sticks (one less than number of boxes) and 5 balls. In a particular arrangement, the 2 sticks divides the balls into 3 buckets or the number of balls that will go into each box.
For example: 
 https://i60.tinypic.com/zvwv1t.png

In the above arrangement 1st stick is at position 2 and 2nd stick is at position 5, this would mean there is 1 ball (ball at position 1) in box 1, 2 (balls at position 3,4) in box 2 and another 2 (balls at position 6,7) in box 3.

We know that number of permutations of x similar items and y similar items is (x+y)! / (x!*y!)

Now lets look at the cases
case d) as both boxes and balls are identical the number of possible arrangements will be 7! / (2!*5!)
case c) both different - 7!
case b) balls different but boxes same - 7! / 2!
case a) boxes different but balls same - 7! / 5!

PS: As it is not clear from the question it is assumed that a box can be empty as well. If not than you can allot 1 ball to each boxes and than do the above on remaining 2 balls and 2 sticks.

Permutation and Combination Conceptual Question

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