Permutations and Combinations
15.0 Division and Distribution of Objects
15.0 Division and Distribution of Objects
The division and distribution of objects with fixed number of objects in each group.
(A) Into groups of unequal sizes:
Let there be $n$ distinct objects divided into $r$ groups of unequal sizes and ${a_1}$ objects belong to first group, ${a_2}$ belong to second group and so on.
Then the number of ways in which the objects can be divided is,
$$\begin{equation} \begin{aligned} { = ^n}{C_{{a_1}}}{.^{n - {a_1}}}{C_{{a_2}}}{.^{n - {a_1} - {a_2}}}{C_{{a_3}}}......{.^{{a_r}}}{C_{{a_r}}} \\ = \frac{{n!}}{{{a_1}!\left( {n - {a_1}} \right)!}}.\frac{{(n - {a_1})!}}{{{a_2}!\left( {n - {a_1} - {a_2}} \right)!}}.\frac{{(n - {a_1} - {a_2})!}}{{{a_3}!\left( {n - {a_1} - {a_2} - {a_3}} \right)!}}.....\frac{{(n - {a_1} - {a_2} - ... - {a_{r - 2}})!}}{{{a_{r - 1}}!\left( {{a_r}} \right)!}}\frac{{{a_r}}}{{{a_r}!\left( 0 \right)!}} \\ = \frac{{n!}}{{{a_1}!{a_2}!{a_3}!....{a_r}!}} \\\end{aligned} \end{equation} $$
and $$n = {a_1} + {a_2} + {a_3} + .... + {a_r}$$
When this $r$ groups have to be distributed to $r$ people, there are $r!$ ways possible. Thus number of ways possible is $$ = {{n!r!} \over {{a_1}!{a_2}!{a_3}!....{a_r}!}}$$
(B) Into groups of equal sizes:
Let there be $m \times n$ objects.
The objects are divided into $n$ groups of $m$ elements each.
Then the number of ways is, $$ = \frac{{(m \times n)!}}{{n!{{(m!)}^n}}}$$
This is because there are $n!$ ways of repetition.
This when distributed among $n$ people is, $$\begin{equation} \begin{aligned} = \frac{{(m \times n)!}}{{n!{{(m!)}^n}}}n! \\ = \frac{{(m \times n)!}}{{{{(m!)}^n}}} \\\end{aligned} \end{equation} $$
Question 29. Consider the following set of coupon numbers, {$1,2,3,4,5,6$}. Find the total numbers of ways in which the set can be
(A) Divided into groups of $a$,$b$ and $c$ with $1$,$2$ and $3$ coupons respectively.
(B) Distributed among $3$ people with one having $1$ coupon, other having $2$ coupons and other having $3$ coupons.
(C) Divided into groups, containing $2$ coupons each.
(D) Distributed to $3$ people, such that everyone gets equal number of coupons.
Solution:
(A) The first group has $1$ coupon, second has $2$ coupons and third has $3$ coupons. There are totally $6$ objects.
Thus, $$\begin{equation} \begin{aligned} = \frac{{6!}}{{1!2!3!}} \\ = 6 \times 5 \times 2 \\ = 60 \\\end{aligned} \end{equation} $$
Group '$a$' may have any one coupon out of $6$, and group '$b$' gets two out of the five, and group '$c$' gets the last three.
Thus possible ways are, $$\begin{equation} \begin{aligned} = {\,^6}{C_1}{.^5}{C_2}{.^3}{C_3} \\ = 6 \times \frac{{5!}}{{2!3!}} \\ = \frac{{6 \times 5 \times 4}}{2} \\ = 60 \\\end{aligned} \end{equation} $$
(B) The number of ways in which it can be distributed is $3!$ times.
So, the number of ways of dividing is, $$\begin{equation} \begin{aligned} = \,\frac{{3!6!}}{{1!2!3!}} \\ = 6 \times 5 \times 4 \times 3 \\ = 360 \\\end{aligned} \end{equation} $$
(C) Now, when dividing into each group having two coupons each is,
| {$1,2$} | {$3,4$} | {$5,6$} |
| {$2,1$} | {$3,4$} | {$5,6$} |
| {$1,2$} | {$4,3$} | {$5,6$} |
| {$2,1$} | {$4,3$} | {$6,5$} |
| {$1,2$} | {$3,4$} | {$6,5$} |
| {$2,1$} | {$3,4$} | {$6,5$} |
| {$1,3$} | {$2,4$} | {$5,6$} |
| {$1,6$} | {$3,5$} | {$2,4$} |
Here, row $1$ to row $6$ i.e. $3!$ are the same and rest are unique.
Thus there are $3!$ repetition for each division.
Thus the number of ways of distribution is, $$\begin{equation} \begin{aligned} = \frac{{6!}}{{2!2!2!3!}} \\ = \frac{{6!}}{{{{(2!)}^3}3!}} \\ = \frac{{6 \times 5 \times 4}}{{2 \times 2 \times 2}} \\ = 3 \times 5 \\ = 15 \\\end{aligned} \end{equation} $$
(D) Distributing the coupons equally among $3$ people can be done in $3!$ ways.
Thus total number of ways is, $$\begin{equation} \begin{aligned} = \frac{{6!3!}}{{(2!)3!}} \\ = \frac{{6 \times 5 \times 4 \times 3}}{{2 \times 2}} \\ = 6 \times 5 \times 3 \\ = 90 \\\end{aligned} \end{equation} $$
