There are twenty-five identical marbles to be divided among four broth

Question

There are twenty-five identical marbles to be divided among four brothers such that each one of them gets no less than three marbles. In how many ways can the marbles be divided among four brothers?

(A) 286
(B) 364
(C) 455
(D) 560
(E) 650

Bunuel · Accepted Answer

The method of (number of groups)^(number of items to distribute) works if the items we distribute are distinct, but in this case, the marbles are identical. If the items are not distinct, we should use the "Stars and Bars" method. For examples, on this check  this post .

"Stars and Bars" method:

Since each brother must get at least 3 marbles, we are essentially distributing only the remaining 25 - 3*4 = 13 marbles.

Imagine the 13 marbles as 13 stars in a row. To divide these stars among 4 brothers, we need 3 bars to create 4 separate sections. For example:

***|****|***|*** would represent that the first brother got 3 chocolates, the second got 4, the third got 3, and the fourth got 3.

||**********|*** would represent that the first brother got 0 chocolates, the second got 0, the third got 10, and the fourth got 3.

****||******|*** would represent that the first brother got 4 chocolates, the second got 0, the third got 6, and the fourth got 3.

So, the problem becomes one of arranging these 13 identical stars and 3 identical bars in a row, which is given by 16!/(13!3!) = 560.

Answer: D

Hope it helps.

firas92 · Answer

Since each brother must get at least 3 marbles, lets first distribute the marbles such that each has 3

This makes us lose 3*4=12 marbles

Now the remaining 13 marbles must be distributed in such a way that any one can get any share

Consider the 13 marbles being separated by 3 separators such that there are four slots in all (representing the four brothers)

*************|||

We can arrange these 16 elements in  \frac{16!}{13!3!}  ways = 560 ways

Alternatively this can be calculated using the formula  (n+r-1)C(r-1)  where  n  identical items are distributed among  r  participants such that any participant may receive any number of items

(13+4-1)C(4-1) = 16C3 = 560

Answer is  (D)

abhishek31 · Answer

Hi  firas92  can you further elaborate on the below point.

"Now the remaining 13 marbles must be distributed in such a way that any one can get any share

Consider the 13 marbles being separated by 3 separators such that there are four slots in all (representing the four brothers)

*************|||

We can arrange these 16 elements in  \frac{16!}{13!3!}  ways = 560 ways"

Thanks.

firas92 · Answer

Hi  abhishek31

That is just a visualization of the formula  (n+r−1)C(r−1)  where  n  identical items are distributed among  r  participants such that any participant may receive any number of items.

In our case, the constraint we are given is that each brother must have at least 3 marbles. How the distribution proceeds after that is not constrained in any way. Which means, once each brother has 3 marbles, we can distribute the remaining in any ratio (13-0-0-0) or (10-1-1-1) or (5-5-2-1) and so on. And remember that there is only 1 way in which we can distribute 3 marbles to each brother because the marbles are all identical.

Now I have represented the 13 marbles as *************

Now let's create 4 slots using separators ||| (each slot representing one brother) - we have to use 3 separators to make 4 slots

____Slot 1___   |   ___Slot 2___   |   ___Slot 3___   |   ___Slot 4___

The way in which the separators are inserted determines the share. (For example  *************||| = 13,0,0,0 ;  ****|****|**|*** = 4,4,2,3 ;  ******|*******|| = 6,7,0,0 ; and so on)

The above are just 3 examples of many cases. So how many ways are there to insert the separators? To calculate this, we just have to find the number of permutations involving 13 identical marbles and 3 identical separators which is given by  \frac{16!}{13!3!}=560ways

Hope this is clear!

singhall · Answer

To more people like me who were wondering what are the stars and bars, here is a complete walk through.
 https://youtu.be/40HxI6Uc00Q

All the best!

rickyric395 · Answer

Hey  Bunuel  can you please provide your 2 cents on this problem. I was thinking along the lines :- first divide 3 marbles to each 4 brother, then divide the remaining 13. Now each marble has 4 choices i.e 1st,2nd,3rd, or 4th brother. So  4^{13} . But obviously its wrong, Can you please help me understand how am I supposed to think in such questions. Also if you could redirect me to similar questions, that would be really helpful.

rickyric395 · Answer

Thanks  Bunuel . It helped a lot.

Jxmes · Answer

can anyone provide insight into why this is incorrect? not really a fan of multiple approaches for special cases, even though it is expedient. The stars and bars makes sense, but would appreciate an explanation on why this does not work. Thanks!

crimson_noise · Answer

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