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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
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
- Q1: How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
- Q2: How many 4-digit numbers are there with no digit repeated?
- Q3: How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
- Q4: Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
- Q6: Find $n$ if ${}^{n–1}P_3 : {}^nP_4 = 1 : 9$.
- Q7(i): Find $r$ if (i) ${}^5P_r = 2 \cdot {}^6P_{r-1}$
- Q7(ii): Find $r$ if (ii) ${}^5P_r = {}^6P_{r-1}$.
- Q8: How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
- Q9(i): How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. (i) 4 letters are used at a time,
- Q9(ii): How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. (ii) all letters are used at a time,
- Q9(iii): How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. (iii) all letters are used but first letter is a vowel?
- Q10: In how many of the distinct permutations of the letters in MISSISSIPPI do the four $I$'s not come together?
- Q11(i): In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S,
- Q11(ii): In how many ways can the letters of the word PERMUTATIONS be arranged if the (ii) vowels are all together,
- Q11(iii): In how many ways can the letters of the word PERMUTATIONS be arranged if the (iii) there are always 4 letters between P and S?
CBSE Solutions for Class 11 Mathematics Permutations and Combinations
Chapters in CBSE - Class 11 Mathematics
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