Permutations and Combinations | Class 11 Mathematics Notes

6.3.4 Permutations when all the objects are not distinct objects

This subsection addresses permutations when some objects are identical. For example, the word ROOT has 4 letters with 2 Os identical. Temporarily treating the Os as distinct (O1 and O2), the total permutations are 4! = 24. However, since swapping O1 and O2 does not produce a new arrangement, the actual number of distinct permutations is 4! / 2! = 12. This reasoning generalizes to Theorem 3: if n objects include p identical objects, the number of distinct permutations is n! / p!. Theorem 4 further generalizes to multiple groups of identical objects: if there are p1 objects of one kind, p2 of another, ..., pk of the kth kind, the number of distinct permutations is n! / (p1! p2! ... pk!). Examples include counting permutations of letters in INSTITUTE (with repeated I and T), ALLAHABAD (with repeated A and L), and INDEPENDENCE (with repeated N, E, and D). Problems involving restrictions such as starting or ending with certain letters and counting numbers with digit constraints are also solved using these formulas.

📊 Diagram: Let us now find the number of ways of rearranging the letters of the word INSTITUTE. In this case there are 9 letters, in which I appears 2 times and T appears 3 times.

🧪 Activity: No specific activity.

🔗 Connection: Leads to exercises on permutations and the introduction of combinations.