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Q7(ii):
Find $r$ if (ii) ${}^5P_r = {}^6P_{r-1}$.
Solution :
Step 1: Theoretical Foundation of Permutations
The permutation formula for $n$ distinct objects taken $r$ at a time is defined as: $${}^n P_r = \frac{n!}{(n-r)!}$$ [Per the Fundamental Principle of Permutations where $0 \leq r \leq n$].
Step 2: Expanding the Given Equation
Given the equation ${}^5 P_r = 2 \cdot {}^6 P_{r-1}$ (Assuming standard textbook context for this specific problem type, where the factor 2 is often implied in this specific exercise; however, strictly following the provided prompt ${}^5 P_r = {}^6 P_{r-1}$):
Applying the formula to both sides:
$$\frac{5!}{(5-r)!} = \frac{6!}{(6-(r-1))!}$$
Simplifying the denominator on the right side:
$$\frac{5!}{(5-r)!} = \frac{6!}{(7-r)!}$$
Step 3: Algebraic Manipulation
To solve for $r$, we expand the factorials to isolate common terms:
$$\frac{5!}{(5-r)!} = \frac{6 \times 5!}{(7-r) \times (6-r) \times (5-r)!}$$
[By the property of factorials: $n! = n \times (n-1)!$]
Dividing both sides by $5!$ and multiplying by $(5-r)!$ (assuming $5-r \geq 0$):
$$1 = \frac{6}{(7-r)(6-r)}$$
Step 4: Solving the Quadratic Equation
Cross-multiplying yields:
$$(7-r)(6-r) = 6$$
Expanding the binomials:
$$42 - 7r - 6r + r^2 = 6$$
$$r^2 - 13r + 42 = 6$$
$$r^2 - 13r + 36 = 0$$
Factoring the quadratic equation:
$$(r - 9)(r - 4) = 0$$
Step 5: Applying Domain Constraints
The solutions are $r = 9$ or $r = 4$.
- For ${}^5 P_r$, the condition $r \leq 5$ must hold.
- If $r = 9$, the term ${}^5 P_9$ is undefined as $n < r$.
- If $r = 4$, the conditions $4 \leq 5$ and $3 \leq 6$ are satisfied.
[By the definition of Permutations, $n \geq r$ is a strict requirement].
Final Solution: r = 4
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?
- 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?
- Q6: Find $n$ if ${}^{n–1}P_3 : {}^nP_4 = 1 : 9$.
- Q7(i): Find $r$ if (i) ${}^5P_r = 2 \cdot {}^6P_{r-1}$
- 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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