In How Many Ways Can One Divide 12 Different Chocolate Bars Into Four GMAT Problem Solving

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!)

“In how many ways can one divide 12 different chocolate bars into four''- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. GMAT quant section explores the candidates’ numerical abilities and their intellectual thinking capacity to solve quantitative problems. The students must answer the question by calculating it with proper mathematical logic. The students must gather enough knowledge regarding mathematical calculations to solve GMAT Problem Solving questions. The GMAT Quant topic in the problem-solving part demands mathematical problems that can be deciphered with excellent quantitative aptitudes. The candidates can enhance their learning by answering questions from the book “501 GMAT Questions”.

Solution and Explanation:

Approach Solution 1:

The problem statement asks to find out:

• The number of ways one can divide 12 distinct chocolate bars into four packs of 3 bars each.

The number of ways 12 different chocolate bars can be split into four distinct 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

Since the packs cannot be distinguishable, it is required to divide the above answer by 4!, therefore, the accurate answer will be

[12!/(3!)^4]/4! = 12!/[4! (3!)^4]

Correct Answer: (C)

Approach Solution 2:

The problem statement asks to find out the number of ways 12 distinct chocolate bars can be divided into four packs of 3 bars each.
The question can be solved in the simplest and quickest way:
Here 4 packs are not different. It is required to divide the answer by 4

Therefore, the number of ways required = 12!/(3!*3!*3!*3!*4!) = 12!/4!(3!)^4

Correct Answer: (C)

Approach Solution 3:

The problem statement asks to find out the number of ways 12 distinct chocolate bars can be divided into four packs of 3 bars each.

I.ABC DEF GHI JKL
II.ACB EDF GHI JKL
III.DEF GHI JKL ABC and so on.

We have to divide the chocolate into four packs of 3 bars each. However, within each set of 3 chocolates, the position of the chocolate is not important [i.e. ABC is the same as BAC]. Therefore, we need to divide 12! by 3! 4 times.

Further, the position of each set among the 4 is not important [i.e. ABC DEF is the same as DEF ABC]. Hence, we need to divide the result by 4!
Therefore, the number of ways required = 12!/(3!*3!*3!*3!*4!) = 12!/4!(3!)^4

Correct Answer: (C)

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