**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 \(=\left( \frac{52!}{{{(13!)}^{4}}\times 4!} \right)\).