In how many ways can one divide twelve different chocolate bars equall

Question

In how many ways can one divide twelve different chocolate bars equally among four boys?

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

dabaobao · Accepted Answer

TL;DR  
Using Anagram Grid: 
AAA BBB CCC DDD 
12!/(3!3!3!3!) = 12!/(3!)^4

Official Solution  
Credit: Veritas Prep

You need to divide 12 chocolate bars among four boys i.e. you have to make four distinct groups. To boy 1, you can give the first chocolate in 12 ways, the second chocolate in 11 ways and the third chocolate in 10 ways. But we don’t want to arrange the chocolates so you can select 3 chocolates for boy 1 in 12*11*10/3! ways (this is equivalent to 12C3 if you follow the formula). Similarly, you can select 3 chocolates for the second boy in 9*8*7/3! ways (i.e. 9C3), for the third boy in 6*5*4/3! ways (i.e. 6C3) and for the fourth boy in 3*2*1/3! ways (i.e. 3C3)

Therefore, you can distribute 12 chocolates among 4 boys equally in (12*11*10/3!) * (9*8*7/3!) * (6*5*4/3!) * (3*2*1/3!) = 12!/(3!*3!*3!*3!) ways

Alternatively, you can visualize putting the 12 chocolates in a row in 12! ways and drawing a line after every three chocolates to demarcate the groups.

OOO l OOO l OOO l OOO

Since the chocolates within the groups are not arranged, we divide 12! by 3! for every group. Since there are 4 groups, number of ways of making 4 distinct groups = 12!/(3!*3!*3!*3!)

ANSWER: C

briantoth6 · Answer

dabaobao  why don't we need to divide the whole thing by 4!?

I was thinking the solution should be 12!/

dabaobao · Answer

We don't divide by 4! since all boys are different, and hence, their order matters.

BrentGMATPrepNow · Answer

Let the 4 boys be: A, B, C and D

Choose 3 chocolate bars for boy A. Since the order in which we can select the 3 bars doesn't matter, we can use combination. 
We can select 3 bars from 12 bars in 12C3 ways. 
12C3 =  (12)(11)(10)/(3)(2)(1)

Choose 3 chocolate bars for boy B. There are now 9 bars remaining. 
We can select 3 bars from 9 bars in 9C3 ways. 
9C3 =  (9)(8)(7)/(3)(2)(1)

Choose 3 chocolate bars for boy C. There are now 6 bars remaining. 
We can select 3 bars from 6 bars in 6C3 ways. 
9C3 =  (6)(5)(4)/(3)(2)(1)

Choose 3 chocolate bars for boy D. There are now 3 bars remaining. 
We can select 3 bars from 3 bars in 3C3 ways. 
9C3 =  (3)(2)(1)/(3)(2)(1)

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus distribute all 12 bars) in  (12)(11)(10)/(3)(2)(1) ] (9)(8)(7)/(3)(2)(1) ] (6)(5)(4)/(3)(2)(1) ] (3)(2)(1)/(3)(2)(1) ]

Simplify product to get: 12!/(3!)⁴

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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dhruva09 · Answer

Bunuel  can you help here, why we can't solve this using star and dash method.
I was getting using that 15!/3!x12!

Bunuel · Answer

Stars and bars does not apply because the chocolate bars are different. That method is used only when the items are identical and only the counts matter. Here, which specific bars each boy gets matters, so the distribution must account for permutations of distinct items, which is why the correct count is based on 12! and not 15!.

dhruva09 · Answer

Thanks for this  Bunuel

In how many ways can one divide twelve different chocolate bars equall

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