Find the number of ways in which 5 identical objects can be distributed among 3 persons if anyone can get any number of objects.
Find the number of ways in which 5 identical objects can be distributed among 3 persons if anyone can get any number of objects.
The correct option is D 21
Let the 5 identical objects be represented by O's . Here all objects can be represented by same symbol zero because they are all identical.
O O O O O
To distribute the objects to 3 persons, we can do it by making 2 partitions so that 3 groups are formed as shown.
For example, by the above partition, we mean 1st person gets 1 object, 2nd person gets 2 objects and 3rd person gets 2 objects.
Similarly, other partitions are also possible like
Here 1st person do not get any object, 2nd person gets 3 objects and 3rd person gets 2 objects.
So, distribution of 5 objects, essentially means to select the places for 2 partitions (each look like I ), in a group of 5 O's and 2 I's
In other words, we have to select 2 I's from a group of 5 I's + 2 0's = 7 objects
This can be done in 7C2 ways.
Generalization:–––––––––––––––––––
If there are n identical objects that are to be distributed to r different groups,
We take n identical objects as n zeroes.
To distribute the above objects to r groups, we need r-1 partitions. Now, considering the partitions as 1's, we have in total n zeroes and r-1 ones' of which we have to select r-1 ones'.
This can be done in n+r−1Cr−1 ways Conclusion: Distribution of n identical objects in r different boxes can be done in n+r−1Cr−1 ways.
21
Let the 5 identical objects be represented by O's . Here all objects can be represented by same symbol zero because they are all identical.
O O O O O
To distribute the objects to 3 persons, we can do it by making 2 partitions so that 3 groups are formed as shown.
For example, by the above partition, we mean 1st person gets 1 object, 2nd person gets 2 objects and 3rd person gets 2 objects.
Similarly, other partitions are also possible like
Here 1st person do not get any object, 2nd person gets 3 objects and 3rd person gets 2 objects.
So, distribution of 5 objects, essentially means to select the places for 2 partitions (each look like I ), in a group of 5 O's and 2 I's
In other words, we have to select 2 I's from a group of 5 I's + 2 0's = 7 objects
This can be done in 7C2 ways.
Generalization:–––––––––––––––––––
If there are n identical objects that are to be distributed to r different groups,
We take n identical objects as n zeroes.
To distribute the above objects to r groups, we need r-1 partitions. Now, considering the partitions as 1's, we have in total n zeroes and r-1 ones' of which we have to select r-1 ones'.
This can be done in n+r−1Cr−1 ways Conclusion: Distribution of n identical objects in r different boxes can be done in n+r−1Cr−1 ways.
