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7 Jan 2007 23:04:31 IST
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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
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10 Jan 2007 19:39:52 IST
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No of ways is = 28 x 28 x 28 Ans.
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Being a genius is 1% luck and 99% perspiration........ |
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16 Jan 2007 14:38:51 IST
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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.
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You never know what is enough till you know what is more than enough.
Titun |
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16 Jan 2007 14:52:28 IST
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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
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19 Jan 2007 13:49:22 IST
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30x29x28
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19 Jan 2007 22:19:24 IST
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for each box minimum no of balls that shud b kept is 1 n maximum is27......
after that..
if we go in decending order ...did any1 noticed a pattern....
if we take 27 in any of the identical box then case formed is...
27 , 0 , 0.......it has jus' got a single case
again with 26 in any of the identical box .....
26 , 1 ,0...... it too has got 1 case...
but if we go on in decending order...
we find after every 2 consecutive cases....that the no. of casas increases....
i.e.for 27 n 26 we have 1 case each
for 25 n 24 we have 2 cases each
for 23 n 22 we have 3 cases each
...n so on...........
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20 Jan 2007 10:36:21 IST
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i got it in two ways:-1)1ball should be there in each compulsorily so rem are 27 balls & 3boxes so no of ways =27+3c27 =29*28/2=406ways 2) partition method 11111111111111111111111111100 no of arrangements = 29!/27!2! =406 ways
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20 Jan 2007 19:53:33 IST
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swordfish #3 result I'm afraid, no student has been able to answer this question with the correct technique.
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20 Jan 2007 19:57:07 IST
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TOTAL NO. OF BALLS=30
TOTAL NO. OF BAOXES = 3
SINCE EACH BALL CAN BE PUT IN ANY OF THE THREE BOXES. SO, THE TOTAL NO. WAYS IN WHICH 30 BALLS CAN BE PUT IN 3 BOXES= 330
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20 Jan 2007 20:46:00 IST
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Let 1 ball be placed in each box then the no of balls left=27
Now these 27 balls can be distributed in the three boxes in the
following no of ways=27C3
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Being a genius is 1% luck and 99% perspiration........ |
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20 Jan 2007 20:47:47 IST
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Sorry I made a mistake
Total=27C3+27C2 + 27C1
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Being a genius is 1% luck and 99% perspiration........ |
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21 Jan 2007 19:58:11 IST
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I agree with NIKUNJ_94's answer
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22 Jan 2007 01:01:54 IST
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CHANDRASEKHAR sir, isnt it 27. I am pretty sure with it till now. Please disclose the answere. I know that every box can be filled with 1 to 27 balls and no matter what it is or which box it is as they all aer identical. so I guessed 27.
Am I wrong? Atlest email me sir.
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Praveen kumar gorakavi
Hyderabad.
gorakavipraveen@gmail.com |
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23 Jan 2007 20:20:06 IST
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The answer is 30-1C3-1=29C2=406
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-ADARSH
NITK Surathkal
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24 Jan 2007 21:17:33 IST
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First give 1ball to each box
The remaining 27 balls can be distributed in the 3 boxes in 27C3 ways
Therefore required no. of ways is 1X27C3=27C3
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25 Jan 2007 11:47:18 IST
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The correct ans is definitely 105
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25 Jan 2007 13:35:26 IST
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vaibhavbright but how an you be so confident, whats you solouition man!!!!!!!!!!
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Praveen kumar gorakavi
Hyderabad.
gorakavipraveen@gmail.com |
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27 Jan 2007 21:04:25 IST
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first let us find out no. of ways of destributing 30 identical balls in 3 different box...(X Y Z)
(27+3-1) C (3-1)
=(29) C (2)
=406 cases
X+Y+Z=27.
(((.note..... X !=Y meansXis not equal to Y)))
IN THIS WE HAVE 3 CASES...
(1) when X=Y=Z.......it will hav only 1 case.....(a)
(2) when X !=Y=z....&....X=Y!=Z......&...X=Z!=Y....
it has....
(1,1,25)
(2,2,23)
...
..
..
&so on till (13,13,1)....
i.e. 13 cases...
&we have 3 such identical boxes......
therefore we have 13 * 3 = 39 cases.......(b)
(3)when X !=Y!=Z
it has( total no. of ways - cases in which atleast 2 are equal)
i.e. 406 -40
=366.....(c)
as we have identical balls therefore we are considering the same cases 6 times ....therefore we will devide...
366 / 6 = 61......(d)
therefore total no of ways are (a)+(b)+(d)...
i.e.1+13+61 = 75......
therefore answere is 75......
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life isn't about waitin' 4 the storm 2 pass by , it's 'bout dancing in rains............
preksha_madan@yahoo.co.in
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27 Jan 2007 21:10:49 IST
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first let us find out no. of ways of destributing 30 identical balls in 3 different box...(X Y Z)
(27+3-1) C (3-1)
=(29) C (2)
=406 cases
X+Y+Z=27.
(((.note..... X !=Y meansXis not equal to Y)))
IN THIS WE HAVE 3 CASES...
(1) when X=Y=Z.......it will hav only 1 case.....(a)
(2) when X !=Y=z....&....X=Y!=Z......&...X=Z!=Y....
it has....
(1,1,25)
(2,2,23)
...
..
..
&so on till (13,13,1)....
i.e. 13 cases...
&we have 3 such identical boxes......
therefore we have 13 * 3 = 39 cases.......(b)
(3)when X !=Y!=Z
it has( total no. of ways - cases in which atleast 2 are equal)
i.e. 406 -40
=366.....(c)
as we have identical balls therefore we are considering the same cases 6 times ....therefore we will devide...
366 / 6 = 61......(d)
therefore total no of ways are
i.e.1+13+61 = 75......
therefore answere is 75......
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life isn't about waitin' 4 the storm 2 pass by , it's 'bout dancing in rains............
preksha_madan@yahoo.co.in
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27 Jan 2007 21:22:00 IST
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n i would like 2 know ur answer vaibhav....
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life isn't about waitin' 4 the storm 2 pass by , it's 'bout dancing in rains............
preksha_madan@yahoo.co.in
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