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Permutation and combination

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Permutation and combination

by parveen110 » Mon Jan 20, 2014 11:42 pm
There are 5 different boxes and 7 identical balls. All the balls are to be distributed in the 5 boxes placed in a row so that any box can recieve any number of balls. In how many ways can these balls be distributed into these boxes?

a.110
b.220
c.330
d.440

OA:C
Last edited by parveen110 on Fri Jan 24, 2014 9:23 am, edited 1 time in total.
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by theCodeToGMAT » Tue Jan 21, 2014 1:08 am
Using Separator MEthod:

Balls = 7
Separators = 4

(7+4)!/7!*4!

= 330
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by GMATGuruNY » Tue Jan 21, 2014 5:08 am
The problem as posted first describes the balls as different but then describes them as identical.
I assume that the intention of the problem is as follows:
There are 5 different boxes and 7 identical balls. All of the balls are to be distributed among the boxes that any box can receive any number of balls. In how many ways can the balls be distributed among the boxes?

a.110
b.220
c.330
d.440

OA:C
The following is called the SEPARATOR method.

Since there are 7 identical balls and 5 boxes, the 7 balls are to be separated into -- at most -- 5 groupings.
Thus, we need 7 balls and 4 separators:
O|O|O|O|OOO

Each arrangement of the elements above represents one way to distribute the 7 balls among 5 boxes A, B, C, D and E:
O|O|O|O|OOO = A gets 1 ball, B gets 1 ball, C gets 1 ball, D gets 1 ball, E gets 3 balls.
OOOOOO||||OO = A gets 5 balls, E gets 2 balls.
OOOOOOOO|||| = A gets all 7 balls.
And so on.

To count all of the possible distributions, we simply need to count the number of ways to arrange the 11 elements above (the 7 identical balls and the 4 identical separators).
The number of ways to arrange 11 elements = 11!.
But when an arrangement includes identical elements, we must divide by the number of ways to arrange the identical elements.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Thus, we must divide by 7! (the number of ways to arrange the 7 identical balls) and by 4! (the number of ways to arrange the 4 identical separators):
11!/(7!4!) = 330.

The correct answer is C.

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