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Marbles and different pockets

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by Mission2012 » Sun Aug 25, 2013 12:10 pm
In how many ways 5 different marbles can be distributes in 3 different pockets so that each pocket has atleast 1 marble?

Ans : 150
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by [email protected] » Sun Aug 25, 2013 3:32 pm
Hi Mission2012,

What were the answer choices for this question? While complex Quant questions will have long-winded "math solutions", sometimes the answer choices can help you to avoid wasting time spent doing lots of math and find a more logical, elegant solution.

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by vipulgoyal » Sun Aug 25, 2013 9:24 pm
let the 5 differant marbles are a,b,c,d,e
and 3 pockets are 1, 2,3

no of cases are
first pocket{(3)number of marbles} * second pocket (1) * third pocket (1) * 3c2 = 60

+

first pocket{(2)number of marbles} * second pocket (2) * third pocket (1) * 3c2 = 90

hence 60+90 = 150 required no of ways

i tried to explain briefly for any queries you are welcome
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by ganeshrkamath » Sun Aug 25, 2013 9:49 pm
Mission2012 wrote:In how many ways 5 different marbles can be distributes in 3 different pockets so that each pocket has atleast 1 marble?

Ans : 150
Let's assume that the second condition (atleast 1 marble in each pocket) is absent.
Now, for each marble, we can select one of the 3 pockets.
So total combinations = 3^5 = 243____________________(A)

Now, let's assume that one of the 3 pockets is absent.
Total combinations = (2^5-2) * 3 = 90____________________(B)

Another case: 2 of the 3 pockets are absent.
Total combinations = 3____________________(C)

Required value = (A) - (B) - (C)
= 150

Cheers
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by vinay1983 » Sun Aug 25, 2013 11:15 pm
vipulgoyal wrote:let the 5 differant marbles are a,b,c,d,e
and 3 pockets are 1, 2,3

no of cases are
first pocket{(3)number of marbles} * second pocket (1) * third pocket (1) * 3c2 = 60

+

first pocket{(2)number of marbles} * second pocket (2) * third pocket (1) * 3c2 = 90

hence 60+90 = 150 required no of ways

i tried to explain briefly for any queries you are welcome
Are you sure this is correct?

3*1*1*3c2=3*3=9
2*2*1*3c2=4*3=12

Either I might have considered your options wrongly or you might have erroneously typed it.
You can, for example never foretell what any one man will do, but you can say with precision what an average number will be up to!
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by vinay1983 » Sun Aug 25, 2013 11:19 pm
Mission2012 wrote:In how many ways 5 different marbles can be distributes in 3 different pockets so that each pocket has atleast 1 marble?

Ans : 150
How different is the above question from this?

In how many ways can 5 boys be allotted 4 different rooms such that none of the rooms is empty and all of the 5 boys are accommodated?

I need solutions for both the types.

Thanks!
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by vipulgoyal » Mon Aug 26, 2013 12:23 am
3*1*1*3c2=3*3=9
2*2*1*3c2=4*3=12

Actully it is

5c3*2c1*1c1*3c2 +
5c2*3c2*1c1*3c2 = 60+90 =150
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by vipulgoyal » Mon Aug 26, 2013 12:35 am
ans to your next q is

In how many ways can 5 boys be allotted 4 different rooms such that none of the rooms is empty and all of the 5 boys are accommodated

5c2*3c1*2c1*1c1*4c3 = 240
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by Mission2012 » Mon Aug 26, 2013 12:57 am
ganeshrkamath wrote:
Mission2012 wrote:In how many ways 5 different marbles can be distributes in 3 different pockets so that each pocket has atleast 1 marble?

Ans : 150
Let's assume that the second condition (atleast 1 marble in each pocket) is absent.
Now, for each marble, we can select one of the 3 pockets.
So total combinations = 3^5 = 243____________________(A)

Now, let's assume that one of the 3 pockets is absent.
Total combinations = (2^5-2) * 3 = 90____________________(B)

Could you please explain B in a little more detail.
i understand that because we have assumed that there are only 2 pockets and none of them should be empty. 2^5 - 2.
Buy why did you multiply by 3
Another case: 2 of the 3 pockets are absent.
Total combinations = 3____________________(C)

Required value = (A) - (B) - (C)
= 150

Cheers
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by Mission2012 » Mon Aug 26, 2013 1:00 am
vipulgoyal wrote:3*1*1*3c2=3*3=9
2*2*1*3c2=4*3=12

Actully it is

5c3*2c1*1c1*3c2 +
5c2*3c2*1c1*3c2 = 60+90 =150
5c3*2c1*1c1*3c2 +
5c2*3c2*1c1*3c2 = 60+90 =150

Why did we multiply both the cases by 3c2?
Could you please explain
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by vipulgoyal » Mon Aug 26, 2013 1:43 am
as given in the solution take three pockets a,b and c
5c3*2c1*1c1*3c2, here i have assinged 3 marbles to a, 1 to b and 1 to c,now the other two cases are
3 to b 1 to a and 1 to c and 3 to c 1 to a and 1 to b. its almost same like number of arrangements of aab = 3!/2!
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by ganeshrkamath » Mon Aug 26, 2013 2:47 am
Mission2012 wrote:
ganeshrkamath wrote:
Mission2012 wrote:In how many ways 5 different marbles can be distributes in 3 different pockets so that each pocket has atleast 1 marble?

Ans : 150
Let's assume that the second condition (atleast 1 marble in each pocket) is absent.
Now, for each marble, we can select one of the 3 pockets.
So total combinations = 3^5 = 243____________________(A)

Now, let's assume that one of the 3 pockets is absent.
Total combinations = (2^5-2) * 3 = 90____________________(B)

Could you please explain B in a little more detail.
i understand that because we have assumed that there are only 2 pockets and none of them should be empty. 2^5 - 2.
Buy why did you multiply by 3
Another case: 2 of the 3 pockets are absent.
Total combinations = 3____________________(C)

Required value = (A) - (B) - (C)
= 150

Cheers
The 3 pockets are unique. So we should consider the possibility of each pocket being empty as 1 unique combination.

The same reason goes for C as well.

Hope this helps.

Cheers
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by Mission2012 » Mon Aug 26, 2013 5:20 am
Got it. thanks a lot
ganeshrkamath wrote:
Mission2012 wrote:
ganeshrkamath wrote:
Mission2012 wrote:In how many ways 5 different marbles can be distributes in 3 different pockets so that each pocket has atleast 1 marble?

Ans : 150
Let's assume that the second condition (atleast 1 marble in each pocket) is absent.
Now, for each marble, we can select one of the 3 pockets.
So total combinations = 3^5 = 243____________________(A)

Now, let's assume that one of the 3 pockets is absent.
Total combinations = (2^5-2) * 3 = 90____________________(B)

Could you please explain B in a little more detail.
i understand that because we have assumed that there are only 2 pockets and none of them should be empty. 2^5 - 2.
Buy why did you multiply by 3
Another case: 2 of the 3 pockets are absent.
Total combinations = 3____________________(C)

Required value = (A) - (B) - (C)
= 150

Cheers
The 3 pockets are unique. So we should consider the possibility of each pocket being empty as 1 unique combination.

The same reason goes for C as well.

Hope this helps.

Cheers
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by Mission2012 » Mon Aug 26, 2013 5:29 am
Got it. Thanks a lot
vipulgoyal wrote:as given in the solution take three pockets a,b and c
5c3*2c1*1c1*3c2, here i have assinged 3 marbles to a, 1 to b and 1 to c,now the other two cases are
3 to b 1 to a and 1 to c and 3 to c 1 to a and 1 to b. its almost same like number of arrangements of aab = 3!/2!
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