Question: In how many ways can 6 chocolates be distributed among 3 children? A child may get any number of chocolates from 0 to 6 and all the chocolates are identical.
A) 21
B) 28
C) 56
D) 112
E) 224
“In how many ways can 6 chocolates be distributed among 3 children”- is a topic that belongs to the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”.
The candidate must have better qualitative aptitudes in order to decipher GMAT Problem Solving questions. GMAT quant helps the candidates to brush up on their mathematical and analytical capabilities. The candidate must pick the valid option by resolving the question with mathematical calculations. The mathematical problems in the problem-solving part of the GMAT Quant topic need to be solved with suitable mathematical knowledge.
Solution and Explanation:
Approach Solution 1:
The problem statement informs that
Given:
- A child may get any number of chocolates from 0 to 6
- all the chocolates are identical
Find Out:
- How many ways can 6 chocolates be distributed among 3 children
The candidate can get the number 6 as the sum of 3 non-negative integers. The integers must be in non-increasing order. The integers are as follows:
{6, 0, 0}, {5, 1, 0}, {4, 2, 0}, {3, 3, 0}, {4, 1, 1}, {3, 2, 1} and {2, 2, 2}
Within each of the sets given above, the candidate can permute the numbers in the following sequence.
Child A gets 6 chocolates, child B gets 0 and child C gets 0 is diverse from child A gets 0, child B gets 6 and child C gets 0. Therefore, the numbers in the above sets can be permuted in the following way:
3!/2! = 3, 3! = 6, 3! = 6, 3!/2! = 3, 3!/2! = 3, 3! = 6, and 3!/3! = 1 way, respectivel
Hence, the number of ways 6 chocolates can be distributed is equal to the sum of 3, 6, 6, 3, 3, 6 and 1. The result of the sum is equal to 28. Thus, there exist 28 ways to distribute 6 chocolates among 3 children.
Hence, option B is the correct answer.
Correct Answer: B
Approach Solution 2:
The problem statement discloses that
Given:
- A child may receive chocolates from 0 to 6
- all the chocolates are identical
Find Out:
- The ways 6 chocolates can be distributed among 3 children
Let us assume that three children get a, b and c chocolates respectively.
That is, the sum of a, b and c is equal to six.
If a is equal to 0, then (b,c) can be (0,6) (1,5) (2,4) (3,3) (4,2) (5,1) (6,0) i.e. 7 ways
If a is equal to 1, then (b,c) can be (0,5) (1,4) (2,3) (3,2) (4,1) (5,0) i.e. 6 ways
If a is equal to 2, then (b,c) can be (0,4) (1,3) (2,2) (3,1) (4,0) i.e. 5 ways
and so on…
Therefore, the total number of ways six chocolates can be distributed among three children will be the sum of the number of ways.
Total ways of distributing 6 chocolates = 7+6+5+4+3+2+1 = 28
Hence option B is the correct answer.
Correct Answer: B
Approach Solution 3:
The problem statement states that a child may receive chocolates from 0 to 6 and all the chocolates are identical. It is required to find out the number of ways 6 chocolates can be distributed among 3 children.
This question can be answered by comparing it to another analogous question: In how many ways can we arrange the letters in IIOOOOOO?
Let's define the relationship between the original question and this question.
First, let's assume the three children are child A, child B and child C.
The arrangement of the 8 letters can be OOIOIOOO
This arrangement basically depicts child A getting 2 chocolates, child B getting 1 chocolate and child C getting 3 chocolates.
Another possible arrangement may be OIOOOOIO
This arrangement depicts child A getting 1 chocolate, child B getting 4 chocolates and child C getting 1 chocolate.
The arrangement can also be OOIIOOOO that represents child A getting 2 chocolates, child B getting 0 chocolate and child C getting 4 chocolates.
Thus the number of ways 6 chocolates can be distributed to 3 children can be determined by doing a permutation of the above = (8!/2! 6!) = 28
Hence, option B is the correct answer.
Correct Answer: B
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