IIT JEE , AIEEE Forum - Mathematics , Algebra - Contest [swordfish #3]: Find ways to distribute identical balls in identical boxes

piyushmalik

important is that the balls are identical and also the boxes.

NUMBER OF WAYS POSSIBLE TO DIVIDE 30 IDENTICAL BALLS TO 3 BOXES IS =

30

C - 1

3

1 IS SUBTRACTED SINCE THERE WILL BE A CASE WHEN A BOX IS EMPTY.

rahul_c

let x,y,z be the no. of balls in each of the boxes.
x+y+z=30
1<=x,y,z<=27
solution =coefficient of a^30 in (a+a^2+a^3 +.....)^3
=30-1C(3-1)=406
HENCE THE BALLS CAN BE DISTRIBUTED IN 406 WAYS

tejal_iitian

tejal_iitian

30c3-3

vaibhavbright

The ans is 56

vaibhavbright

The ans is 56

malay

the answer is 65

malay

the solution is:
one assumption:first box contains the minimum no. of balls and the third box contains the maximum no. of balls(this doesn,t change the answer since all boxes are identical)
firstly fill the boxes with one ball each
we are left with 27 balls which can be distributed in second and third box in 13 ways.
next, fill the boxes with two balls in each
we are left with 24 balls which can be distributed in 12 ways.
hence the answer is 13+12+10+9+7+6+4+3+1=65

Any doubt???

malay

I left one case (10,10,10)
so the answer is 66(rest 65 cases are discussed in earlier post)

sampathkaushik

This can be answered easily by using the multinomial theorem

each box has to have atleast one ball; and hence a max of 28 balls
and there are 3 boxes;

hence ans is the coeff of x³⁰ in the expression
{x + x² + x³ + x⁴ + ....... x²⁸ }³
or coeff of x²⁷ in the expression
{1 + x² + x³ + ..... x²⁷}³

which is nothing but ²⁹C₂
which is equal to 406

uday_sravan

The total no. of ways in which 30 balls can be distributed among 3 boxes is 30+3-1C3-1=496.

No. of ways any of the boxes can get zero no. of balls is 6.

So, the required answer is 496-6=490

10904him

No of ways is = 28 x 28 x 28 Ans.

titun

First, let us put one ball in each of the 3 boxes. So, now no box remains empty & we have to just distribute 27 balls in 3 identical boxes.

The no. of ways by which this can be done = (The no. of non-negative integral solutions of the eqn. a + b + c = 27 ) = [(3+ 27-1)C27]/3 = 29C27 = 406

Here, I have made an assumption that the case when x no. of balls are in the 1st box, y no. of ball are in the 2nd box and the remaining balls in the 3rd box & the case when y no. of balls are in the 1st box, x no. of ball are in the 2nd box and the remaining balls in the 3rd box are distict cases and both of them are taken into account.

Strictly speaking, these two are not distinct cases as the boxes are identical.

So, actual no. of cases = 406/3

But since this is not an integer, I have made the above assumption.

ariyam66

The reqd. no. of ways = no. of positive integral solutions of

p+q+r = 30

Let p = m + 1, q = n + 1, r = k + 1

Therefore, m+n+k = 27

Hence, no. of integral solns of m+n+k=27 is 29C2

But since the boxes are identical, total no. of ways = (29C2)/3

Ask & Discuss Questions with Community & Experts