56.

here's how.

a minimum of a ball per box. let box B and C have 1 ball each. then box A has 28. keep on decreasing a ball from box A and figure the combinations in boxes B and C. the total is 1+1+2+2+3+3+4+4+5+5+6+6+7+7.

for first box, any one ball can be placed by 30 ways

for second box, one ball can be placed by 29 ways

for third box, one ball can be placed by 28 ways

remaining balls = 27

each ball can be placed by 3 ways

total ways = 30 X 29 X 28 X

No of ways of arranging 30 balls in 3 identical boxes = 27

3ⁿ -2

each ball has 3 options but no box should be empty (3³⁰ -2)/3

guys the ans is 56 only by mistake, actually the ans is 75

sorry for my previous post, the answer is 75, wll post solution later

Wow!! just look at hte number of views for this problem

Ok people here is my approach to solve this problem:

We are suppose to find the number of ways to distribute 30 identical balls in 3 Identical boxes none empty

Now consider dis dummy problem no. positive integral of solutions of x+y+z = 30

Answer of this dummy prob. is ²⁹C₂ = 406.

But wait not so fast 406 will definitely not be the answer for our main question because in the dummy problem x, y, z were distinct.

For us x, y, z are identical.

so what to do???

Out of 406 solutions 1 corresponds to the solution {10,10,10} which has been counted only once.

Now we are remaining with other 405 solutions.

Out of theses 405 solutions there are cases like {5,5,20}, {5,20,5}, {20,5,5,} in general two same solutions which have been counted

3 times. (let number of such unique solutions be i )

Also there solutions like {12,8,10},  {12,10,8},  {8,10,12}, {8,12,10},.... in general all three different which have benn counted

3! = 6times  (let number of such unique solutions be j)

so 3i + 6j = 405

Now to find no. of solutions in which any two of x, y, z are same

i.e. to find no. of  non-negative solutions of 2g + h =27  .................(no box should be empty so give them each 1 ball does not matter as                                                                                                                        all balls are identical)

which are clearly 13   ......................... g can vary from 0 to 13      but g = 9 is the set {10,10,10} which has been taken care of.

 so i=13 .............no. of sol. in which two are equal

now  3*13 +  6j = 405                             hence j = 61

Hence the total number of solutions are

all same + two same + no same (all different ) = 1 + i + j = 1 + 13 + 61 = 75

Hence the answer is 75!!!

Hats of! :-) i guess the admin who wud have made this post around some 2 years wud have finally got the correct explanation with an apt solution.

yup, but this one is easier one!!

case  with (look down)                      no. of such possible cases

28 balls in one box & rest in others=     1
27 in one =                                                  1
26 in one =                                                  2
25 in one =                                                   2
24 in one =                                                   3
23 in one =                                                    3
22 in one =                                                    4
21 in one =                                                   4
20 in one =                                                    5
19 in one =                                                    5
18  in one =                                                   6
17 in one =                                                    6
16 in one =                                                     7
15 in one =                                                     7

14 in one=           8-1                                    7      ( one counted in above case)

13 in one =        8-3                                     5  (similar reason)

12 in one=         9-5                                    4

11 in one =       9-7                                     2

10 in one  =     10-9                                   1    stop, no more cases possible

these all add up to 75, this one is easy i think.....

it is actually a counting roblem...not a permutation problem...

the answer will be  30C₃  i.e 30! / 27!3!

 sorry, but here's it gain, i want to know the right answer, so here's my trial

case  with (look down)                      no. of such possible cases

28 balls in one box & rest in others=     1 
27 in one =                                                  1 
26 in one =                                                  2 
25 in one =                                                   2 
24 in one =                                                   3 
23 in one =                                                    3 
22 in one =                                                    4 
21 in one =                                                   4
20 in one =                                                    5 
19 in one =                                                    5 
18  in one =                                                   6 
17 in one =                                                    6 
16 in one =                                                     7 
15 in one =                                                     7

14 in one=           8-1                                    7      ( one counted in above case)

13 in one =        8-3                                     5  (similar reason)

12 in one=         9-5                                    4

11 in one =       9-7                                     2

10 in one  =     10-9                                   1    stop, no more cases possible

these all add up to 75, this one is easy i think.....

27C3