Distribution of Identical objects into Distinct Groups.

How to Distribute identical objects into Distinct Groups
Tutorial on Distribution of Identical Objects into Distinct Groups.

Distribution of ‘n’ identical objects into ‘m’ numbers of distinct groups.

PODCAST ON DISTRIBUTION OF OBJECTS INTO GROUPS

Let us consider distribution of n identical objects to m numbers of distinct groups.

let $\,x_i \,$ is the number of identical object objects in $i^{th}\, $ group.

Therefore $x_1\,+\,x_2\,+\,x_3……….\,+\,x_m\,=\,n\,\,\, ;\,\,\, where\,\, 0\leq x_i\leq n \,,\,\,\, \forall i=1,2,3,……..m $

Hence number of non negative integral solution of equation will be the number ways in which n identical objects can be distributed into m numbers of groups .

The number of solutions is Coefficient of $x^n\,$ in the expansion of $ (1+x+x^2+………+x^n)^m $

=Coefficient of $ x^n $ in the expansion of $ (1-x)^{-m}\,\, =\, {n+m-1 \choose n} $

Hence number of distribution = $ {n+m-1 \choose n} $

Distribution of n identical objects into m numbers of groups such none of the group remain empty.

Total number of solutions will be positive integral solution of equation of equation

$ x_1+x_2+x_3…………..+x_m\,=\,n \,\,\, 1\leq x_i \leq n \,\, \forall i=1,2,3……n $

let $ x_i\,=\,y_i\,+1\,\,\,\, therefore\,\, y_i\geq 0 \,\,\, \forall x_i \geq 1. $

Therefore number of distribution will be number of non negative solution of equation

$ y_1+y_2+y_3…………..+y_m\,=\,n\,-\,m \,\,\, 0\leq y_i \leq n-1 \,\, \forall i=1,2,3……n $

Hence Number of solutions are $ {(n-m)+m-1 \choose (n-m)}\,=\,{n\,-\,1 \choose n-m} $

Applications of distribution of identical objects

Question 1. Find the number of ways in which 10 identical blankets can be distributed to 5 beggars.

Answer :::: Here n =10 , m = 5 Hence $ {14 \choose 10} $

To study the distribution of Distinct objects into groups click here

Question 2 . Find coefficient of $x^5$ in the expansion of $(\,x^2\,-\,x\,+\,1\,)^4$.

Answer :::: General term in the expansion of $(\,x^2\,-\,x\,+\,1\,)^4$ is

$T_{(r_1,\,r_2,\,r_3)}=\frac{4!}{{r_1}!{r_2}!{r_3}!}\,(x^2)^{r_1}\,(-x)^{r_2}\,(1)^{r_3} $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=(-1)^{r_2}\,\frac{4!}{{r_1}!{r_2}!{r_3}!}\,x^{(2{r_1}\,+\,{r_2})} $

For Coefficient of $x^5$

$2{r_1}\,+\,r_2\,=\,5$ and $\,\,r_1\,+\,r_2\,+\,r_3\,=\,4$

Hence $(\,r_1\,,r_2\,,r_3\,)$ = ( 1 , 3 , 0 ) and ( 2 ,1 ,1 )

Therefore Coefficient of $x^5$ is = $\,\frac{4!}{1!\,3!\,0!}\,(-1)^3\,+\,\frac{4!}{2!\,1!\,1!}\,(-1)^1 $ $\\$ $= -6$

Question 3. Find number of ways in which from n or more number of identical object of 1 type , n or more number of identical object of 2 nd type and …. n or more number of identical object of m th type , exactly n can be selected.

Answer :::: Here also number of ways will be number of non negative solution of equation

$$x_1+x_2+x_3……..x_m=n$$

Where $x_i$ denotes number of i th objects taken in a selection.

And number of selection will be $${n+m-1 \choose n }$$

Question 4: Find the number of ways 4 persons from 12 sitting around a circular table be selected such that no two of them are neghibour.

Answer :::: let persons selected are A,B,C,D and $X_1\,,\,X_2\,,\,X_3\,,\,X_4$ be number of persons between A,B ; B,C ; C,D ; D,A respectively.

Hence $1\le X_i\le 8\,\,\, \forall i=1,2,3,4$

Number of selection will be number of integral solution of equation $$X_1+X_2+X_3+X_4=8$$

$$Let\,\,\,X_i=Y_i+1\,\,\,\,where\,\,\, 0\le Y_i \le 7 $$

Hence Number of selection is number of non negative solution of equation $$Y_1+Y_2+Y_3+Y_4=4$$

$$\therefore Solutions ={7 \choose 4}=35 $$

Question 5. Find the number of ways in which 12 people can be arranged around the circular table so that particular 4 persons A, B, C and D are not together.

Answer :::: Number of ways in which 8 other than A , B , C and D can be arranged around the circular table is 7! And between these 8 . A , B , C and D be arranged in $^8P_4$.

Hence total number of ways$$ =( 7! )(^8P_4)$$.

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