# Permutations and Combinations – Division of Objects into Groups

## Permutations and Combinations – Division of Objects into Groups

Division of items into groups of unequal size:

• Number of ways in which (m + n)! distinct items can be divided into two unequal groups containing m and n items is (m + n)! / (m!. n!).
• The number of ways in which (m + n + p) items can be divided into unequal groups containing m, n, p items is $$^{m+n+p}{{C}_{m}}{{.}^{n+p}}{{C}_{n}}=\frac{(m+n+p)!}{m!\times n!\times p!}$$.
• The number of ways to distribute (m + n + p) items among 3 persons in the groups containing m, n and p items is = (No. of ways to divide) x (No. of groups)! $$=\frac{(m+n+p)!}{m!\times n!\times p!}\times 3!$$.

Division of objects into groups of equal size:

• The number of ways in which mn distinct objects can be divided equally into m group each containing n objects and the order of the groups is not important is $$\left[ \frac{(mn)!}{{{(n!)}^{m}}} \right]\times \frac{1}{m!}$$.
• The number of ways in which mn different items can be divided equally into m group each containing n objects and the order of group is important is   $$\left( \frac{(mn)!}{{{(n!)}^{m}}}\times \frac{1}{m!} \right)\times m!=\frac{(mn)!}{{{(n!)}^{m}}}$$.

Examples 1: In how many ways can a pack of 52 cards be divided equally among four players in order?

Solution: The number of ways in which mn different items can be divided equally into m group each containing n objects and the order of group is important is $$=\frac{(mn)!}{{{(n!)}^{m}}}$$ .

Here 52 cards are to be divided into four equal groups and the order of the groups is important so required number of ways $$=\left( \frac{52!}{{{(13!)}^{4}}\times 4!} \right)\times 4!=\frac{52!}{{{(13!)}^{4}}}$$.

Example 2: In how many can a pack of 52 cards be divided equally into four group?

Solution: The number of ways in which mn distinct objects can be divided equally into m group each containing n objects and the order of the groups is not important is $$\left[ \frac{(mn)!}{{{(n!)}^{m}}} \right]\times \frac{1}{m!}$$.

Here order is not important the total number of ways is $$=\left( \frac{52!}{{{(13!)}^{4}}\times 4!} \right)$$.