There are 15 chocolates be distributed among 3 children subject to condition that child can take any number of chocolates. Find the required number of ways to do this if chocolates are identical.
There are 15 chocolates be distributed among 3 children subject to condition that child can take any number of chocolates. Find the required number of ways to do this if chocolates are identical.
The correct option is D
Let the number of chocolates given to each child be A, B and C
So, A + B + C = 15
Let the 15 chocolates be represented by 15 O's. Here all chocolates can be represented by same symbol zero because they are all identical.
O O O O O O O........15 times
Now these 15 chocolates are to be distributed to 3 children.
To distribute the 15 chocolates to 3 persons, we can do it by making 2 partitions so that 3 groups are formed as shown.
O|OO|O00000000000
For example, by the above partition, we mean 1st child gets 1 chocolate, 2nd child gets 2 chocolates and 3rd child gets 12 chocolates.
Similarly, other partitions are also possible.
So, distribution of 15 chocolates, essentially means to select the places for 2 partitions (each look like I), in a group of 15 O's and 2 I's
In other words, we have to select places for 2 I's out of 17 places.
This can be done in 15+3−1C3−1 = 17C2 ways.
Generalization:
If there are n identical objects that are to be distributed to r different groups such that each group gets at least 1 object,
We take n identical objects as n zeroes.
As each group must get at least 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 at least 1 object, can be done in n−1Cr−1 ways.
Let the number of chocolates given to each child be A, B and C
So, A + B + C = 15
Let the 15 chocolates be represented by 15 O's. Here all chocolates can be represented by same symbol zero because they are all identical.
O O O O O O O........15 times
Now these 15 chocolates are to be distributed to 3 children.
To distribute the 15 chocolates to 3 persons, we can do it by making 2 partitions so that 3 groups are formed as shown.
O|OO|O00000000000
For example, by the above partition, we mean 1st child gets 1 chocolate, 2nd child gets 2 chocolates and 3rd child gets 12 chocolates.
Similarly, other partitions are also possible.
So, distribution of 15 chocolates, essentially means to select the places for 2 partitions (each look like I), in a group of 15 O's and 2 I's
In other words, we have to select places for 2 I's out of 17 places.
This can be done in 15+3−1C3−1 = 17C2 ways.
Generalization:
If there are n identical objects that are to be distributed to r different groups such that each group gets at least 1 object,
We take n identical objects as n zeroes.
As each group must get at least 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 at least 1 object, can be done in n−1Cr−1 ways.
