In how many ways can 6 chocolates be distributed among 3 children? A c

Question

In how many ways can 6 chocolates be distributed among 3 children?  A child may get any number of chocolates from  1 to 6  and all the chocolates are identical .

A) 10
B) 15
C) 21
D) 28
E) 56

SOURCE :  https://www.GMATinsight.com

amanvermagmat · Accepted Answer

Hello

The formula of identical objects DOES work here. If we have to distribute N identical objects among R distinct groups such that one or more people might get none of the objects, then the formula is = (N+R-1) C (R-1) (selecting R-1 objects out of N+R-1 objects)

BUT if N identical objects have to be distributed among R distinct groups such that everyone should get at least one object, then the formula is = (N-1) C (R-1) (selecting R-1 objects out of N-1 objects)

So in this question, since 6 identical chocolates have to be distributed among 3 distinct people, but everyone should get at least one, we will apply the second formula = (6-1) C (3-1) = 5C2 = 10, which is our answer

Bunuel · Answer

vitaliyGMAT · Answer

Hi

# of chocolates distributed to each child - x_1, x_2, x_3 , where  x_i > 0

We have non-empty set:

x_1 + x_2 + x_3 = 6

We need to convert it into  x_i >=0  substituting each  x_i  with  y_i = x_i - 1 .

x_i = y_i +1 :

y_1 + y_2 + y_3 = 3

_{3+3-1}C_3 = _5C_3 = \frac{5*4}{2} = 10

Answer A

testcracker · Answer

\ since a child must get at least 1 chocolate, lets distribute 1 chocolate to each child first, and thus we are left with 3 chocolates to redistribute now we are in business since chocolates are identical, the remaining 3 chocolates can be distributed among 3 children as follows 5! _____ 3! 2! = 10, the answer hope this helps thanks cheers, and do consider some kudos, guys

Archit3110 · Answer

formula to use here we can take case that child may get 0 chocolate 
n-1Cr-1 ; n=6 . r= 3 
5c2; 10
IMO A

exc4libur · Answer

given: 6 identical chocs, 3 different kids, at least 1 each;

k_1+k_2+k_3=6…(k_1'+1)+(k_2+1)+(k_3+1)=6…k_1'+k_2+k_3=3 
 C(n+r-1,r-1)=(3+3-1,3-1)=\frac{5!}{2!3!}=10

Answer (A)

GMATinsight · Answer

The detailed solution to the above problem using   two methods   is explained in the attached video

https://www.youtube.com/watch?v=Ooc96NawyIY

ScottTargetTestPrep · Answer

Let the children be A, B, and C. So A can get 1, B can get 1 and C can get 4 chocolates. Of course, this is different from A gets 4, B 1, and C 1, or, A gets 1, B 4, and C 1.

In the calculations below, we will show how 3 positive integers can sum to 6 and the number of ways the 3 numbers can be rearranged among A, B, and C (for example, the first calculation below describes the distribution of the 6 chocolates mentioned above):

1 + 1 + 4 = 6

3!/2! = 3 ways

1 + 2 + 3 = 6

3! = 6 ways

2 + 2 + 2 = 6

3!/3! = 1 way

Therefore, there are a total of 3 + 6 + 1 = 10 ways that 6 chocolates can be distributed to 3 children.

Answer: A

joon259 · Answer

One issue I have with the wording is the fact that if each child must get a minimum of 1 chocolate, then it's impossible for a child to get 6 chocolates if there are only 6 that are being distributed.

Krunaal · Answer

That's a good point, and I feel that the sentence could've been better framed. However, if you look at it as range: 0 < c < 7, where c is no. of chocolates => while it is  necessary  that each child gets at least 1 chocolate, but at the same time it just implies that the no. of chocolates a child gets is less than 7, it is  not necessary  that a child gets 5 or 6 chocolates.

PSKhore · Answer

Thanks for the formula!

We can also add a slight twist to the formula (N+R-1) C (R-1), if we don't remember the new one on time.

Distribute one chocolate to each child, and now we are left with 3 chocolates and 3 children. So, we can apply the original formula (N+R-1) C (R-1) because now we do have the possibility of someone having 0 chocolates from the remaining chocolates.

5C2 = 10

In how many ways can 6 chocolates be distributed among 3 children? A c

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GMAT Problem Solving (PS) Questions

In how many ways can 6 chocolates be distributed among 3 children? A c

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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