Home » Ask & Discuss » Mathematics. » Algebra « Back to Discussion

Algebra

Administrator

Joined:	19 Oct 2006
Post:	1558

18 Nov 2006 03:14:59 IST

0 People liked this

125

9507

Contest [swordfish #3]: Find ways to distribute identical balls in identical boxes

None

Find the no. of ways of distributing 30 identical balls in the three identical boxes, such that no box remains empty.

Share this article on:

Comments (125)

Go to Page...

uday_sravan

New kid on the Block

Joined: 3 Jan 2007

Posts: 16

7 Jan 2007 23:04:31 IST

Like 0 people liked this

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

Hot goIITian

Joined: 29 Dec 2006

Posts: 191

10 Jan 2007 19:39:52 IST

Like 0 people liked this

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

Titun

Forum Expert

Joined: 23 Dec 2006

Posts: 374

16 Jan 2007 14:38:51 IST

Like 0 people liked this

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.

Das Mukerjee

Cool goIITian

Joined: 23 Dec 2006

Posts: 50

16 Jan 2007 14:52:28 IST

Like 0 people liked this

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

ijaz17

New kid on the Block

Joined: 19 Jan 2007

Posts: 1

19 Jan 2007 13:49:22 IST

Like 0 people liked this

30x29x28

preksha madan

Cool goIITian

Joined: 17 Jan 2007

Posts: 35

19 Jan 2007 22:19:24 IST

Like 0 people liked this

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...........

trachion_me

New kid on the Block

Joined: 20 Jan 2007

Posts: 1

20 Jan 2007 10:36:21 IST

Like 0 people liked this

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

chandrasekhar

Forum Expert

Joined: 12 Nov 2006

Posts: 25

20 Jan 2007 19:53:33 IST

Like 0 people liked this

swordfish #3 result I'm afraid, no student has been able to answer this question with the correct technique.

rachitaw

New kid on the Block

Joined: 20 Jan 2007

Posts: 1

20 Jan 2007 19:57:07 IST

Like 0 people liked this

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

10904him

Hot goIITian

Joined: 29 Dec 2006

Posts: 191

20 Jan 2007 20:46:00 IST

Like 0 people liked this

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

10904him

Hot goIITian

Joined: 29 Dec 2006

Posts: 191

20 Jan 2007 20:47:47 IST

Like 0 people liked this

Sorry I made a mistake
Total=27C3+27C2 + 27C1

zuberahmed

Cool goIITian

Joined: 18 Dec 2006

Posts: 38

21 Jan 2007 19:58:11 IST

Like 0 people liked this

I agree with NIKUNJ_94's answer

gorakavipraveen

Hot goIITian

Joined: 5 Dec 2006

Posts: 169

22 Jan 2007 01:01:54 IST

Like 0 people liked this

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.

CyBorG

Forum Expert

Joined: 6 Jan 2007

Posts: 706

23 Jan 2007 20:20:06 IST

Like 0 people liked this

The answer is ^30-1C_3-1=²⁹C₂=406

tejasvi2389

Hot goIITian

Joined: 24 Nov 2006

Posts: 133

24 Jan 2007 21:17:33 IST

Like 0 people liked this

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

vaibhavbright

Cool goIITian

Joined: 27 Dec 2006

Posts: 89

25 Jan 2007 11:47:18 IST

Like 1 people liked this

The correct ans is definitely

105

gorakavipraveen

Hot goIITian

Joined: 5 Dec 2006

Posts: 169

25 Jan 2007 13:35:26 IST

Like 0 people liked this

vaibhavbright but how an you be so confident, whats you solouition man!!!!!!!!!!

preksha madan

Cool goIITian

Joined: 17 Jan 2007

Posts: 35

27 Jan 2007 21:04:25 IST

Like 1 people liked this

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......

preksha madan

Cool goIITian

Joined: 17 Jan 2007

Posts: 35

27 Jan 2007 21:10:49 IST

Like 1 people liked this

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......

preksha madan

Cool goIITian

Joined: 17 Jan 2007

Posts: 35

27 Jan 2007 21:22:00 IST

Like 0 people liked this

n i would like 2 know ur answer vaibhav....

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team