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16 Dec 2006 11:28:12 IST
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There is only one way to do that.
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16 Dec 2006 12:10:21 IST
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ans:30C 3 =30!/27!*3!=4060 george
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16 Dec 2006 13:35:14 IST
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1010
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16 Dec 2006 14:14:32 IST
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1
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17 Dec 2006 12:22:17 IST
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n-1Cr-1 =no empty box is allowed. so every ball occupies 1 space for sure 29x14=406
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goutham |
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17 Dec 2006 17:49:46 IST
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FOR ALL SUCH PROBLEMS ANSWER IS n-1Cr-1 THUS ANS IS 29C2 . BUT BOXES R IDENTICAL SO ANS IS (29C2)/3
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17 Dec 2006 19:59:15 IST
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the solutions given by all the above people are incorrect, i will give you the method, try to rectify the solution and get the correct answer, first of all consider that the boxes are different, then the no of ways of distributing 30 identical balls into 3 boxes so that no box remains is 30-1 C 3-1 which will be equal to 29 C2 which is equal to 406, in these 406 cases there are certain cases in which all the boxes contain unequal balls ex 1,2,27 each of these has been considered 6 times, but the boxes are indentical therefore only one sixth of these should be taken, similarly there are certain cases in which two boxes have equal no of ball ex 2,2,26 which have been considered 3 only one third of these case must be taken , and there is one case with equal no of balls if you proceed in this manner get the right answer, the correct answer is 75
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17 Dec 2006 21:58:21 IST
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answer to the question is  n+r-1(C)r-1
therefore
30+3-1(C)3-1=496
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anupam |
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17 Dec 2006 22:20:30 IST
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req.ans is soln. of equation x1+x2+x3=30. x3(1+x+x2+x3+-----------------+x27)3 (3+27-1)c 27==406 ans==406
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19 Dec 2006 12:19:29 IST
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The solution involves finding the co-efficient of X^30 in the expression,
(x + x^2 + x^3 + .... +X^30)^3
because the problem is essentially finding the solutions to equation,
a+b+c = 30 ; a,b,c belong to N.
The coefficient comes out to be 336.
But in the question, the boxes are also identical, so, this should be divided by 6 (3!)
, so the answer is 336/6 = 56.
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29 Dec 2006 13:18:43 IST
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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.
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29 Dec 2006 15:16:53 IST
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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
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The yardstick of human intelligence is the ability to overcome the last fallacy |
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29 Dec 2006 16:54:55 IST
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30p3
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29 Dec 2006 16:57:36 IST
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30c3-3
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30 Dec 2006 13:08:09 IST
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The ans is 56
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30 Dec 2006 13:09:12 IST
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The ans is 56
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Thinktank - the official forum of IIT Roorkee is back : http://people.iitr.ernet.in/thinktank/ |
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30 Dec 2006 14:23:07 IST
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the answer is 65
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Imagination is more important than knowledge
-------Albert Einsetein |
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30 Dec 2006 16:46:41 IST
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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???
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Imagination is more important than knowledge
-------Albert Einsetein |
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31 Dec 2006 10:24:17 IST
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I left one case (10,10,10)
so the answer is 66(rest 65 cases are discussed in earlier post)
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Imagination is more important than knowledge
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7 Jan 2007 01:10:58 IST
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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
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