In how many ways can one divide 12 different chocolate bars into four

Question

In how many ways can one divide 12 different chocolate bars into four packs of 3 bars each?

A) 12!/3!4!
B) 12!/(4!)^2
C) 12!/4!(3!)^4
D) 12!/4!(4!)^3
E) 12!/4!(3!4!)

dabaobao · Answer

TL;DR Anagram Grid: AAA AAA AAA AAA (not same as AAA BBB CCC DDD) 12!/((3!3!3!3!)*4!) = 12!/4!(3!)^4 Official Solution Credit: Veritas Prep Similar to Q: https://gmatclub.com/forum/in-how-many- ... l#p2278052 What do you think will the answer be here? Will it be the same as above? No. Here the 4 packs are not distinct. You need to divide the answer you obtained above by 4! (similar to the simple example with just 4 chocolates we saw above). In this case, the required number of ways = 12!/(3!*3!*3!*3!*4!) Since the groups are not distinct here, your answer changes. When the question says that you need to make n groups/bundles/teams that are not distinct, you need to divide by (n!). If the groups/bundles/teams are distinct then you do not divide by (n!). ANSWER: C

puneetb · Answer

Do you not think that arrangement within the Stacks would matter provided the Chocolates are different, and we are not just making teams?

IanStewart · Answer

I agree - it's really not clear what the question means. Is there a first stack, a second stack, a third and a fourth? Does it matter what chocolate is at the top of each stack and which is on the bottom? I'd just be guessing how to answer a question worded this way.

The interpretation I'd find most reasonable is the one you suggest - if we are making 'stacks' (and not just groups or sets), I'd assume there's some reason the question is using that word, and that it matters what is on the top, in the middle and at the bottom of each stack. But that's not the interpretation used in the 'official solution'.

dabaobao · Answer

Agreed! I just modified the Q from "stacks" to "packs" to remove the ambiguity.

ScottTargetTestPrep · Answer

The number of ways 12 different chocolate bars can be divided into four different packs of 3 bars each is:

12C3 x 9C3 x 6C3 x 3C3 = 12!/(9!3!) x 9!/(6!3!) x 6!/(3!3!) x 1 = 12!/(3!3!3!3!) = 12!/(3!)^4

However, since here the packs are indistinguishable, we have to divide the above answer by 4!, and therefore, the correct answer is:

/4! = 12!/ 
 
Answer: C

aritrar4 · Answer

ScottTargetTestPrep   VeritasKarishma

I understood the first part of the explanation to arrive at 12C3 x 9C3 x 6C3 x 3C3 = 12!/(3!)^4

Could you please explain the second part? 
"However, since here the packs are indistinguishable, we have to divide the above answer by 4!"

I understand dividing the total combinations by N! if N items are identical in a list of different items (e.g. arranging letters in MISSISSIPPI = 11!/(4!4!2!)), but I'm having trouble to understand how the 4 identical packs would lead to that logic. Since the packs are different entities from the bars themselves.

Thanks !

KarishmaB · Answer

12! of the numerator involves all arrangements such as

I.
ABC DEF GHI JKL

II.
ACB EDF GHI JKL 
 
III.
DEF GHI JKL ABC

etc

Now here is the point - should I and II be two different ways? No. You need groups of 3 chocolates. The chocolates inside the groups should not be arranged. So you divide by 3! four times.
Now think - should I and III be different ways? I still have the same 4 groups. Whether DEF is kept first or ABC is irrelevant. I have the same 4 groups irrespective of how the groups are kept. Hence I should divide by 4! too.

aritrar4 · Answer

This makes sense now. Thank you!

Posted from my mobile device

ScottTargetTestPrep · Answer

While the chocolate bars are distinguishable, once it is determined which three bars will be packed together, the order of the packs are not important. Here's what I mean: suppose (xyz) and (uvw) are two packs of chocolate. The number 12!/(3!)^4 counts the packings (xyz), (uvw), * , * and (uvw), (xyz), * , * as different ways to pack the 12 bars (* denotes the remaining packs, irrelevant); however, these two packings actually correspond to a single way to divide the 12 bars into 4 packs. In fact, this same packing appears many times within the 12!/(3!)^4 ways to pack the chocolates; we can have (uvw), * , (xyz), *; or *, *, (uvw), (xyz); or (uvw), *, *, (xyz) etc. The exact number of duplicate counts can be found by the number of ways to arrange the four packs once it's determined which three bars will go into each pack and that number is 4! (since there are four packs). That's why we need to divide 12!/(3!)^4 by 4! to get the actual number of ways to divide the 12 bars into packs of 3.

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In how many ways can one divide 12 different chocolate bars into four

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In how many ways can one divide 12 different chocolate bars into four

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