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Q5:
From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?

Solution :

Step 1: Identifying the Mathematical Framework

The problem requires the selection of two distinct positions—Chairman and Vice Chairman—from a set of $n = 8$ distinct individuals. Because the roles are distinct (the order of selection matters; choosing Person A as Chairman and Person B as Vice Chairman is different from choosing Person B as Chairman and Person A as Vice Chairman), this is a Permutation problem.

Mathematically, we are calculating the number of arrangements of $r=2$ objects chosen from a total of $n=8$ objects [Per the Fundamental Counting Principle and Permutation definition].

Step 2: Mathematical Formulation

The number of ways to arrange $r$ items out of $n$ distinct items is given by the formula:

${}^nP_r = \frac{n!}{(n-r)!}$

Where:

  • $n = 8$ (Total number of persons available)
  • $r = 2$ (Number of positions to be filled)

Step 3: Execution of Calculation

Substitute the given variables into the formula:

${}^8P_2 = \frac{8!}{(8-2)!} = \frac{8!}{6!}$

Expand the factorial terms to simplify the expression [By the definition of factorial $n! = n \times (n-1) \times (n-2) \times \dots \times 1$]:

${}^8P_2 = \frac{8 \times 7 \times 6!}{6!}$

Cancel out the common $6!$ term:

${}^8P_2 = 8 \times 7 = 56$

Visual Representation of Selection Logic

Chairman (8 choices) Vice Chair (7 choices) Total = 8 × 7 = 56 Ways

Step 4: Logical Verification via Multiplication Principle

Alternatively, using the Fundamental Principle of Counting:

  • Step A: Selection of a Chairman. Since there are 8 persons, there are 8 ways to fill this position.
  • Step B: Selection of a Vice Chairman. Since one person cannot hold two positions, there are $(8 - 1) = 7$ remaining candidates.

By the Multiplication Principle [Per the Fundamental Counting Principle], the total number of ways to perform both steps is $8 \times 7 = 56$.

Final Solution: The total number of ways to choose a chairman and a vice chairman is 56.


More Questions from Class 11 Mathematics Permutations and Combinations EXERCISE 6.3


CBSE Solutions for Class 11 Mathematics Permutations and Combinations


Chapters in CBSE - Class 11 Mathematics


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