# Division and distribution of objects

(with fixed number of objects in each group)

## Into groups of unequal size
(different number of objects in each group):

(a) Number of ways in which n distinct objects can be
divided into r unequal groups containing a_{1} objects in the first
group, a_{2} objects in the second group and so on

= =

Here a_{1 }+ a_{2 }+ a_{3 }+ …
+a_{r} = n.

(b)
Number of ways in which n distinct objects can be distributed among r persons
such that first person gets a_{1} objects, 2^{nd} person gets a_{2}
objects…,r^{th} person gets a_{r} objects = .

### Explanation:

Let us divide the
task into two parts . In the first part, we divide the objects into groups. In
the second part, these r groups can be assigned to r persons in r! ways. .

## Into groups of equal size
(each group containing same number of objects):

(a) Number of ways in
which m´n distinct objects
can be divided equally into n groups (unmarked) =

(b) Number of ways in which m´ n different objects can be
distributed equally among n persons (or numbered groups) = (number of ways of
dividing into groups)´(number of groups)!
=

## Example.1

If out of 50 players, 5 teams of 10 players
each have to be formed this becomes a question on grouping and thus required
number of ways to form such teams is

## Derangement

Let S = {1, 2, 3, …. ,n}, then a function f
from S to S known as derangement if f is a bijective function and f(i) > i for any i ∈ S.

In
other words rearrangement of objects such that no one goes to its original
place is called derangement

If
'n' things are arranged in a row, the number of ways in which they can be
deranged so that none of them occupies its original place is

= n!and it
is denoted by D(n)

Note:

The above result can be obtained by using inclusion
exclusion principle. For this you can refer to problem –13 on page 29.

## Example.2

Suppose 4 letters are taken out of 4 different
envelopes. In how many ways, can they be reinserted in the envelopes so that no
letter goes in to its original envelope ?.

### Solution

Using the formula for the number of derangements that
are possible out of 4 letters in 4 envelopes, we get the number of ways as :