Let's assume that the second condition (atleast 1 marble in each pocket) is absent.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
Are you sure this is correct?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
How different is the above question from this?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
ganeshrkamath wrote:Let's assume that the second condition (atleast 1 marble in each pocket) is absent.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
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
5c3*2c1*1c1*3c2 +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
The 3 pockets are unique. So we should consider the possibility of each pocket being empty as 1 unique combination.Mission2012 wrote:ganeshrkamath wrote:Let's assume that the second condition (atleast 1 marble in each pocket) is absent.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
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
ganeshrkamath wrote:The 3 pockets are unique. So we should consider the possibility of each pocket being empty as 1 unique combination.Mission2012 wrote:ganeshrkamath wrote:Let's assume that the second condition (atleast 1 marble in each pocket) is absent.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
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 same reason goes for C as well.
Hope this helps.
Cheers
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!
