Distribution of things concept is used to find the number of ways of distributing n distinct objects in r distinct boxes. From this concept, questions are frequently asked in JEE and other competitive examinations. In this article, we discuss three cases of distribution of things.
In the first case, we discuss distribution of distinct things in distinct boxes when empty boxes are allowed. In the second case, we consider distribution of distinct things in distinct boxes when empty boxes are not allowed. In the third case, we consider identical things to be distributed in distinct boxes when empty boxes are allowed. While learning Permutations and Combinations, students are recommended to understand distribution of things thoroughly so that they can easily solve related problems.
Formula
Case 1: Empty boxes allowed
The number of ways of distributing n distinct things in r distinct boxes so that each box is filled with 0 or more things (empty boxes allowed) = rn.
Consider that we have n distinct objects t1, t2, t3…tn and r distinct boxes. The number of ways the object t1 can be distributed is r. Similarly, t2 can also be distributed in r boxes in r ways.
The object tn can be distributed in r boxes in r ways.
So the number of ways of distributing n distinct things in r distinct boxes = r×r×…r(n times) = rn.
Case 2: Empty boxes not allowed
The number of ways of distributing n distinct things in r distinct boxes so that each box is filled with at least one thing (empty boxes not allowed)
= rn – rC1(r – 1)n + rC2(r – 2)n – …+ (-1)r-1. rCr-1(1)n
Case 3: Identical things in distinct boxes
The number of ways of distributing n identical things in r distinct boxes so that each box is filled with 0 or more things (empty boxes are allowed) = n+r-1Cr-1
Also Read
JEE Main Permutations and Combinations Previous Year Questions
Principle of Inclusion & Exclusion & Derangement
Solved Examples
Example 1:
Find the number of ways of putting 5 rings in 4 boxes so that each box is filled with 0 or more rings.
a. 45
b. 54
c. 20
d. None of these
Solution:
Given number of rings, n = 5
Number of boxes, r = 4
First ring can be kept in any of the 4 boxes.
Second ring can also be put into any of the 4 boxes.
Similarly the fifth ring has 4 choices.
So the number of ways of putting 5 rings in 4 boxes = 4×4×4×4×4
= 45
Hence, option a is the answer.
Example 2:
The total number of ways in which 5 shirts of different colours can be distributed among 3 persons so that each person gets at least one shirt is:
a. 180
b. 150
c. 200
d. None of these
Solution:
The number of ways of distributing n distinct things in r distinct boxes so that each box is filled with at least one thing = rn – rC1(r – 1)n + rC2(r – 2)n – …+ (-1)r-1. rCr-1(1)n
Here number of shirts, n = 5, number of persons, r = 3
The total number of ways in which 5 shirts of different colours can be distributed among 3 persons so that each person gets at least one shirt is = 35 – 3C1(3 – 1)5 + 3C2(3 – 2)5
= 243 – 3(25) + 3(1)
= 246 – 96
= 150
Hence, option b is the answer.
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Frequently Asked Questions
In how many ways can you distribute n distinct things in r distinct boxes such that each box is filled with 0 or more things?
The number of ways of distributing n distinct things in r distinct boxes so that each box is filled with 0 or more things = rn.
When do we use nCr?
When the order does not matter, we use nCr.
When do we use nPr?
nPr is used when order matters.
In how many ways can you distribute n identical things in r distinct boxes so that each box is filled with 0 or more things?
The number of ways of distribution = n+r-1Cr-1.
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