 # Formation of Groups

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 learn to divide objects in groups or 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 one player. Let A, B, C, 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!)

### 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:

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 in two groups such that each group contains 2 objects each?

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 two, three and five objects?

a. 2435

b. 2520

c. 2000

d. None of these

Solution:

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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