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Algebra
18 Nov 2006 03:14:59 IST
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simply_ayush
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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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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
n+r-1
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.
(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.
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???
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???
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
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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