Find the number of ways in which 7 identical objects can be distributed among 3 persons if each gets atleast one object.
Find the number of ways in which 7 identical objects can be distributed among 3 persons if each gets atleast one object.
The correct option is D 15
Let the 7 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 O O
Condition is given that each person must get atleast 1 object. So, out of 7 objects, first let us give 3 objects to 3 persons so that each person gets atleast one object. As the objects are identical, number of ways to do this is 1.
Number of remaining objects are 7-3 = 4
Now these 4 identical objects are to be distributed to 3 persons, such that anyone can get any number of objects(as each have 1 object already)
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 1 object.
Similarly, other partitions are also possible.
So, distribution of 4 objects, essentially means to select the places for 2 partitions (each look like I ), in a group of 4 O's and 2 I's
In other words, we have to select 2 I's from a group of 4 I's + 2 0's = 6 objects
This can be done in 6C2 ways.
Generalization:–––––––––––––––––––
If there are n identical objects that are to be distributed to r different groups such that each group gets atleast 1 object,
We take n identical objects as n zeroes.
As each group must get atleast 1 object, we first give r objects to r different groups.
Number of remaining objects = n-r
To distribute these n-r objects to r groups, we need r-1 partitions. Now, considering the partitions as 1's, we have in total n-r zeroes and r-1 ones' of which we have to select r-1 ones'.
This can be done in (n−r)+(r−1)Cr−1=n−1Cr−1 ways
Conclusion:–––––––––––––– Distribution of n identical objects in r different boxes such that each box gets atleast 1 object, can be done in n−1Cr−1 ways.
15
Let the 7 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 O O
Condition is given that each person must get atleast 1 object. So, out of 7 objects, first let us give 3 objects to 3 persons so that each person gets atleast one object. As the objects are identical, number of ways to do this is 1.
Number of remaining objects are 7-3 = 4
Now these 4 identical objects are to be distributed to 3 persons, such that anyone can get any number of objects(as each have 1 object already)
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 1 object.
Similarly, other partitions are also possible.
So, distribution of 4 objects, essentially means to select the places for 2 partitions (each look like I ), in a group of 4 O's and 2 I's
In other words, we have to select 2 I's from a group of 4 I's + 2 0's = 6 objects
This can be done in 6C2 ways.
Generalization:–––––––––––––––––––
If there are n identical objects that are to be distributed to r different groups such that each group gets atleast 1 object,
We take n identical objects as n zeroes.
As each group must get atleast 1 object, we first give r objects to r different groups.
Number of remaining objects = n-r
To distribute these n-r objects to r groups, we need r-1 partitions. Now, considering the partitions as 1's, we have in total n-r zeroes and r-1 ones' of which we have to select r-1 ones'.
This can be done in (n−r)+(r−1)Cr−1=n−1Cr−1 ways
Conclusion:–––––––––––––– Distribution of n identical objects in r different boxes such that each box gets atleast 1 object, can be done in n−1Cr−1 ways.
