The formation of groups concept in permutation and combination is used to find the number of ways n distinct objects can be divided into r groups, whose sizes are known. We use permutation for the arrangement of objects in a specific order. Combination is used if some objects are to be arranged in such a way that the order of objects is not important. In this article, we will learn to divide objects into groups, and how to distribute these objects between different people.
Consider that we have 4 players. We want to form 2 teams in which one team has 3 players and the other team has 1 player. Let A, B, C, and D be the players. We can form the team as shown in the table below.
| Team 1 | Team 2 |
| A, B, C | D |
| A, C, D | B |
| A, B, D | C |
| B, C, D | A |
The number of ways of forming teams = 4C3 = 4
Formula
1. Number of ways of dividing m+n (where m ≠ n) things into two unequal groups of size m and n is m+nCm × nCn = (m+n)!/m!n!.
2. Number of ways of dividing (m+n+p) (where m ≠ n ≠ p) things into 3 unequal groups of size m, n, p is given by (m+n+p)!/m!n!p!.
3. Number of ways of dividing 2n things into 2 equal groups of size n each is 2nCn×nCn/2!
= (2n)!/2!(n!.n!)
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Solved Examples
Example 1:
Find the number of ways of distributing 8 different packets among Ram, Srini and Tina such that Ram gets 2 packets, Srini gets 1, and Tina gets 5 packets.
a. 168
b. 186
c. 120
d. None of these
Solution:
The number of ways of dividing (m+n+p) (where m ≠ n ≠ p) things into 3 unequal groups of size m, n, p is = (m+n+p)!/m!n!p!.
Here, m = 2, n = 1 and p = 5
So, the number of ways = (8!/2!.1!.5!)×1
= 168
Hence, option a is the answer.
Example 2:
In how many ways can 4 objects be divided into two groups such that each group contains 2 objects?
a. 3
b. 6
c. 12
d. None of these
Solution:
Number of ways = 4!/2!2!2!
= 24/8
= 3
Hence, option a is the answer.
Example 3:
In how many ways can 10 different objects be distributed in 3 persons in such a way that they get 2, 3 and 5 objects?
a. 2435
b. 2520
c. 2000
d. None of these
Solution:
The number of ways of dividing (m+n+p) (where m ≠ n ≠ p) things into 3 unequal groups of size m, n, p is = (m+n+p)!/m!n!p!.
So, the number of ways = 10!/2!3!5!
= 2520
Hence, option b is the answer.
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Frequently Asked Questions
Give the formula for permutation.
Permutation is given by the formula, nPr = (n!)/(n-r)!
Give the formula for combination.
Combination is given by the formula, nCr = (n!) /(r! (n-r)!).
In how many ways can we divide m+n items into two unequal groups of size m and n, (m ≠ n)?
The number of ways of dividing m+n (m ≠ n) items into two unequal groups of size m and n = (m+n)!/m!n!.
When do we use permutations?
We use permutations when the order of arrangement matters.
In how many ways can we divide 2n things into 2 equal groups?
The number of ways of dividing 2n things into 2 equal groups with size n each is 2nCn×nCn/2! = (2n)!/2!(n!.n!).
When do we use combinations?
We use combinations when the number of possible groups is to be found, and the order is not needed.
Give an example of a combination.
Selecting 3 children from a group of 9 children is an example of a combination.
What is the relation between combination and permutation?
Combination and permutation are related by the formula nCr = nPr/r!.
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