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Counting Problem

Expert replies
by nysnowboard » Wed Mar 10, 2010 1:01 am
Ok, so this isn't a GMAT question per se, but I don't see how it couldn't theoretically appear on a test in the future.

How many ways can we place 7 identical balls into 8 separate (but
distinguishable) boxes?

Now to be fair, multiple choice answers would help us to the solution but there could also be trick answers so I am trying to solve this problem without "assistance" from answer choices (plus there are no multiple choice questions lol).

[spoiler]The solution, as given from an outside source, says it boils down to a simple combination with repetition allowed problem. So C(n+k-1, k)

If n = 8 and k = 7

(8+7-1)!/7!(8+7-1-7)!= 14!/7!7! = total number of ways of placing 7 identical balls into 8 distinct boxes[/spoiler]

Solution is above but I have some questions.

For starters, is this even fair game for the GMAT?

Secondly, how do we determine n and k in this situation? I can't wrap my head around it....

I tried to suppose that instead of boxes and balls, it was, say, the set of ABCDEFGHIIIIIII (8 distinct letters and 7 identical)
which at least made it easier to conceptualize, it's a combination of 15 elements with 7 repeating items and then I realized, I have NO clue of what is being selected and what it is being selected from...

If this is useless then sorry for wasting your time and I will ignore combinations with repetition for GMAT purposes at least =).
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Source: — Problem Solving |

by sanju09 » Wed Mar 10, 2010 3:16 am
nysnowboard wrote:Ok, so this isn't a GMAT question per se, but I don't see how it couldn't theoretically appear on a test in the future.

How many ways can we place 7 identical balls into 8 separate (but
distinguishable) boxes?

Now to be fair, multiple choice answers would help us to the solution but there could also be trick answers so I am trying to solve this problem without "assistance" from answer choices (plus there are no multiple choice questions lol).

[spoiler]The solution, as given from an outside source, says it boils down to a simple combination with repetition allowed problem. So C(n+k-1, k)

If n = 8 and k = 7

(8+7-1)!/7!(8+7-1-7)!= 14!/7!7! = total number of ways of placing 7 identical balls into 8 distinct boxes[/spoiler]

Solution is above but I have some questions.

For starters, is this even fair game for the GMAT?

Secondly, how do we determine n and k in this situation? I can't wrap my head around it....

I tried to suppose that instead of boxes and balls, it was, say, the set of ABCDEFGHIIIIIII (8 distinct letters and 7 identical)
which at least made it easier to conceptualize, it's a combination of 15 elements with 7 repeating items and then I realized, I have NO clue of what is being selected and what it is being selected from...

If this is useless then sorry for wasting your time and I will ignore combinations with repetition for GMAT purposes at least =).
Combinations with Repetition are dogged by coming across at a set of items, and selecting a subset though permitting repetition. As such, Combinations with Repetition are a superset of Combinations, as perceived in the following example:

Combinations with Repetition of {A B C D} choose 2, are:

{A A}, {A B}, {A C}, {A D}, {B B}, {B C}, {B D}, {C C}, {C D}, {D D}

The C(n + k - 1, k) formula best fits here, when n = 4 and k = 2

C(n + k - 1, k) = 5C2 = 10.

I am not sure who to call starters in GMAT, but if anybody is eying a score that’s 720+, such concepts are value to treasure.

Moderators are requested to move the problem to the right bin with a reminder.
The mind is everything. What you think you become. –Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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by nysnowboard » Wed Mar 10, 2010 4:15 am
Ah ok... after re-reading your explanation I think I get the idea... I came across examples like yours but couldn't quite get it, but your superset/subset explanation made sense.


So for the original problem, the superset is the 8 boxes and the subset is the 7 balls; 8 choose 7. And it makes sense. LoL... I guess I will have to drill this into my head before I get confused again.

Also, I assumed this was the proper place for this post, but if it isn't please direct me toward the right forum section so I can rectify the oversight in future posts.

Thank you again, and I somehow figured you'd be the first to answer this lol =)
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by sanju09 » Wed Mar 10, 2010 4:54 am
nysnowboard wrote:Ah ok... after re-reading your explanation I think I get the idea... I came across examples like yours but couldn't quite get it, but your superset/subset explanation made sense.


So for the original problem, the superset is the 8 boxes and the subset is the 7 balls; 8 choose 7. And it makes sense. LoL... I guess I will have to drill this into my head before I get confused again.

Also, I assumed this was the proper place for this post, but if it isn't please direct me toward the right forum section so I can rectify the oversight in future posts.

Thank you again, and I somehow figured you'd be the first to answer this lol =)
If a mathematical doubt is in neither PS nor DS form of GMAT, then the doubt may be posted directly to the main forum called GMAT Math forum. The PS and DS are the sub forums of the same designed for its purpose.

You sound very loll, why don’t you use the emoticons to express yourself better. Keep pouring in :roll:
The mind is everything. What you think you become. –Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Join the discussion