What is the number of ways in which 20 identical marbles can be distri

Question

What is the number of ways in which 20 identical marbles can be distributed among five children, such that each child gets at least 3 marbles?

(A) 125
(B) 126
(C) 127
(D) 128
(E) 129

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Nidzo · Accepted Answer

Another approach using the more common GMAT combination formula:

As each of the 5 children get 3 marbles, there will be 5 marbles left over. 
The different ways in which 5 marbles can be distributed are:

5(one child gets all 5):  As each child could get 5, there are  5  ways in which to distribute 5.

4-1 (one child gets 4 and one child gets 1):  
Number of ways of selecting 2 children from 5:  \frac{5!}{2!3!} = 10 . 
Number of ways of in which one can distribute the marbles to the 2 children selected:  2 . 
Total ways in which 2 kids from the 5 get 4 and 1 extra marble:  10*2=20

3-2 (one child gets 3 and one child gets 2):  This is almost an identical repeat of above, and thus has  20  ways of occurring.

3-1-1 (one child gets 2 and two children get 1):  
Number of ways of selecting 3 children from 5:  \frac{5!}{3!2!} = 10 . 
Number of ways of in which one can distribute the marbles to the 3 children selected:  \frac{ 3!}{1!2!} = 3.  
Total ways in which 2 kids from the 5 get 1 extra marble and 1 kid gets 3 extra marbles:  10*3=30 .

2-2-1 (two children get 2 and one child gets 1):  The number of ways will be identical to the above. There are  30  ways in which this can happen.

2-1-1-1 (one child gets 2 and three get 2):  
Number of ways of selecting 4 children from 5:  \frac{5!}{4!1!} = 5 . 
Number of ways of in which one can distribute the marbles to the 4 children selected:  \frac{4!}{1!3! }= 4 . 
Total ways in which 2 kids from the 5 get 1 extra marble and 1 kid gets 3 extra marbles:  5*4=20 .

1-1-1-1-1 (each child gets 1):  There is only  1  way in which this occurs.

In total there are:  5 + 20 + 20 + 30 + 30 + 20 + 1 = 126

Answer B

sanjitscorps18 · Answer

Essentially this problems asks a + b + c + d + e = 20 where each of the children (a, b, c, d, e) has more than or equal to 3 marbles.
If we distribute 3 marbles each to the children we are left with 5 marbles. Now we have to distribute 5 marbles to 5 children and each of them can have 0 marbles. So the revised equation becomes a' + b' + c' + d' + e' = 5

To do this we have the formula n+r-1Cr-1
To ways = 5+5-1C5-1 = 9C4 = 126 ways

Option B

SaquibHGMATWhiz · Answer

A very interesting one where you know the formula then you can solve it easily.

However, it is not a common type and hence may not be covered in various courses. So I'll discuss this with two different approaches.

• The condition here is that they must get at least 3 and thus we need 15 marbles to distribute and make sure the condition is satisfied.
 
o This means we now have 5 marbles to distribute among 5 children and now they can get 0 (As 3 already given
o There are two different approaches that one can follow here.
  One could be to count scenarios manually such as (5, 0, 0, 0, 0); (4, 1, 0, 0, 0); (3, 2, 0, 0, 0); (3, 1, 1, 0, 0) (2, 2, 1, 0, 0) (2, 1, 1, 1, 0) (1, 1, 1, 1, 1)
 The problem with this is two folds. One, it is a bit tedious and 2nd one is that there's a chance that you will miss out on cases. 
 How to make sure that we do not miss cases?
 Start with bigger numbers and then go in descending order. So here first one (5, 0, 0, 0, 0) and then we get (4, 1, 0, 0, 0).
 Now after this we can say that the maximum one can get is 3 and then 2 so we have (3, 2, 0, 0, 0) and after this, we see if we can keep 3 as the highest and split 2 further. We can do that in (3, 1, 1, 0, 0). 
Similar way we keep on getting the different possibilities.

One may wonder what about cases such as (0, 5, 0, 0, 0). Well, those scenarios we will get by arranging these numbers. 
  (5, 0, 0, 0, 0) =  \frac{{5!}}{{4! × 1!}}  = 5
 (4, 1, 0, 0, 0)=  \frac{{5!}}{{1! × 1! × 3!}}  = 20
 (3, 2, 0, 0, 0)=  \frac{{5!}}{{1! × 1! × 3!}}  = 20
 (3, 1, 1, 0, 0)=  \frac{{5!}}{{1! × 2! × 2!}}  = 30
 (2, 2, 1, 0, 0)=  \frac{{5!}}{{2! × 1! × 2!}}  = 30
 (2, 1,1, 1, 0)=  \frac{{5!}}{{1! × 3! × 1!}}  = 20
 (1, 1, 1, 1, 1)=  \frac{{5!}}{{5! × 1! × 2!}}  = 1

Thus if we add all of them then we get 126. Option B.

Making a separate post for the second method as it is relatively non-conventional concept

SaquibHGMATWhiz · Answer

Now the second approach is what is given in the first post.

However, I am explaining the logic behind the same.

So, we have established that 5 children now need to get 5 marbles such that some of them can get 0 marbles.
Thus we can say that 1st, 2nd, 3rd, 4th, and 5th children can get A, B, C, D, and E numbers of marbles respectively.
So we have A + B + C +D +E = 5 
The solution to this equation comes from the following concept.

Think about there are 5 marbles there as • • • • • 
Now what we will do is to put 4 different barricades to separate these into 5 parts. Understand that to make 5 regions we need only 4 barricades. Let's denote them as | | | |

The significance of these brackets is that,
the area left to the first barricade is for the first child, 
the area between the first and second barricade is for the second child,
the area between the second and the third barricade is for the third child, 
the area between the third and 4th barricade is for the 4th child, and 
the area to the right of the 4th barricade is for the 5th child

Now, all we need to do is to arrange these 5 marbles and 4 barricades together and we can do that in  \frac{{9!}}{{4! × 5!}} 
 Now how does this arrangement decide the number of ways? Let's understand that a little.
We have • • • • • | | | | and this is one possible arrangement which means all the five marbles gone to the first children. 
Let's say another arrangement is • •| | •| • • | - this suggests that the first child gets 2, the second one gets 0, the third one gets 1, 4th one gets 2 and the 5th child gets 0.

So, in a nutshell, we need to know how many marbles and how many barricades are there.
If the number of marbles are 'm' and the number of people is 'p' then the answer is  \frac{{(m + p - 1!)}}{{m! × (n-1)!}} 
This is same as  ^{m+p - 1} C m  or  ^{m+p - 1} C p-1

Oppenheimer1945 · Answer

Lets distribute 3 to 5.(15)
Remaining=5
5+4C4=15C4

9C4=6.7.8.9/24=6.7.3=6*21=126

bumpbot · Answer

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What is the number of ways in which 20 identical marbles can be distri

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