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Permutations and Combinations

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by Rahul@gurome » Sat Aug 07, 2010 5:36 am
aarati wrote:In how many ways 5 different balls can be arranged in to 3 different boxes so that no box remains empty?


help me in solving these..............
Following cases are possible:

(1) 1 ball can be placed in 2 boxes each and 3 balls can be placed in the 3rd box. There are 3 ways in which we can do so.
This can be done in 3 * [5C1 * 4C1 * 3C3] = 3 * [5 * 4 * 1] = 3 * 20 = 60 ways

(2) 2 balls can be placed in 2 boxes each and 1 ball can be placed in the 3rd box. There are 3 ways in which we can do so.
This can be done in 3 * [5C2 * 3C2 * 1C1] = 3 * [10 * 3 * 1] = 90 ways

Total no. of ways = 60 + 90 = 150 ways.
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by muralithe1 » Sun Aug 08, 2010 2:50 pm
hey rahul.
PLease correct me if i am wrong here..

1) 1 ball can be placed in 2 boxes each and 3 balls can be placed in the 3rd box. There are 3 ways in which we can do so.
This can be done in 3 * [5C1 * 4C1 * 3C3] = 3 * [5 * 4 * 1] = 3 * 20 = 60 ways

WHy can first fill the box which can be filled with 3 balls and take the rest two boxes of 1 ball each.
then in that case the result will like below.

3 * (5C3 * 2C1 * 1C1) = 60....

Am I missing something here....

I guess u might need to divide the total by box arrangement as well (3,1,1 - 1,1,3 - 1,3,1 )

Not quite sure..Experts plz correct me here...and show me direction.,
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by Rahul@gurome » Sun Aug 08, 2010 5:53 pm
muralithe1 wrote:hey rahul.
PLease correct me if i am wrong here..

WHy can first fill the box which can be filled with 3 balls and take the rest two boxes of 1 ball each.
then in that case the result will like below.

3 * (5C3 * 2C1 * 1C1) = 60....

Am I missing something here....

I guess u might need to divide the total by box arrangement as well (3,1,1 - 1,1,3 - 1,3,1 )

Not quite sure..Experts plz correct me here...and show me direction.,
You are correct, and that's why I have taken 3 arrangements for case 1. Let me elaborate.

Case 1: 1 balls each in 2 of the boxes and 3 balls in the 3rd box

Box 1: 1, Box 2: 1, Box 3: 3; No. of ways of doing this = 5C1 * 4C1 * 3C3 = 5*4*1 = 20
Box 1: 1, Box 2: 3, Box 3: 1; No. of ways of doing this = 5C1 * 4C3 * 1C1 = 5*4*1 = 20
Box 1: 3, Box 2: 1, Box 3: 1; No. of ways of doing this = 5C3 * 2C1 * 1C1 = 10*2*1 = 20

So, the total no. of ways we can arrange the balls for case 1 = 20 + 20 + 20 = 60 ways

Similarly, we make 3 arrangements for case 2.

Does that answer your question?
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by aarati » Sun Aug 08, 2010 11:36 pm
none of these answers are not correct.... the OA is 720.i don't how to solve it explain me how to do it ........
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by selango » Sun Aug 08, 2010 11:40 pm
aarati,

What is the source of this question?
--Anand--
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by sheelanadh » Tue Aug 10, 2010 2:58 am
aarati wrote:In how many ways 5 different balls can be arranged in to 3 different boxes so that no box remains empty?


help me in solving these..............

(1) 1 ball can be placed in 2 boxes each and 3 balls can be placed in the 3rd box. There are 3 ways in which we can do so.
This can be done in 3 * [5C1 * 4C1 * 3C3] = 3 * [5 * 4 * 1] = 3 * 20 = 60 ways

(2) 2 balls can be placed in 2 boxes each and 1 ball can be placed in the 3rd box. There are 3 ways in which we can do so.
This can be done in 3 * [5C2 * 3C2 * 1C1] = 3 * [10 * 3 * 1] = 90 ways

Total no. of ways = 60 + 90 = 150 ways


please check your OA once?
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by aarati » Tue Aug 10, 2010 3:22 am
selango wrote:aarati,

What is the source of this question?

this is from local institutes.......


please help me to solve it........
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by aarati » Wed Aug 11, 2010 9:27 pm
please help me to solve it.........
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