Permutations and Their Theorems

Data Science and A.I. Lecture Series

Author: Bindeshwar Singh Kushwaha

Institute: PostNetwork Academy

Theorem 1: Permutations of Distinct Objects

The number of permutations of n different objects taken r at a time is:

\[ P(n, r) = \frac{n!}{(n-r)!} \]

Explanation:

• First position: n choices.
• Second position: n-1 choices.
• Continue until the r-th position.

Example: Arranging 3 objects out of 5.

\[ P(5, 3) = \frac{5!}{(5-3)!} = 60 \]

Theorem 2: Permutations with Repetition Allowed

The number of permutations of n different objects taken r at a time, with repetition allowed, is:

\[ n^r \]

Explanation:

• Each position has n choices.
• Total ways: n × n × … × n (r times).

Example: Arranging 2 objects from 3 with repetition.

\[ 3^2 = 9 \]

Theorem 3: Permutations of Objects with Some Identical

The number of permutations of n objects, where p objects are identical, is:

\[ \frac{n!}{p!} \]

Example: Arranging the letters of “ROOT” (2 identical O‘s).

\[ \frac{4!}{2!} = 12 \]

Theorem 4: General Case for Non-Distinct Objects

The number of permutations of n objects, where p₁, p₂, …, p_k are identical, is:

\[ \frac{n!}{p_1! \cdot p_2! \cdot … \cdot p_k!} \]

Example: Arranging the letters of “ALLAHABAD” (4 A‘s, 2 L‘s).

\[ \frac{9!}{4! \cdot 2!} = 7560 \]

Permutations When All Objects Are Not Distinct

Scenario: Some objects are identical, e.g., arranging letters in “ROOT”.

Method:

• Temporarily treat identical objects as distinct.
• Divide total permutations by the factorial of identical objects.

Formula:

\[ \text{Permutations} = \frac{n!}{p_1! \cdot p_2! \cdot … \cdot p_k!} \]

Example: Letters of “INSTITUTE” (2 I‘s, 3 T‘s).

\[ \frac{9!}{2! \cdot 3!} = 30240 \]

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