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Q7(i):
Find $r$ if (i) ${}^5P_r = 2 \cdot {}^6P_{r-1}$

Solution :

Step 1: Theoretical Foundation

The permutation formula is defined as ${}^nP_r = \frac{n!}{(n-r)!}$, where $0 \leq r \leq n$ [Per the fundamental principle of permutations].

Step 2: Substitution into the Given Equation

Given equation: ${}^5P_r = 2 \cdot {}^6P_{r-1}$

Expanding both sides using the permutation formula:

$\frac{5!}{(5-r)!} = 2 \cdot \frac{6!}{(6-(r-1))!}$

$\frac{5!}{(5-r)!} = 2 \cdot \frac{6!}{(7-r)!}$

Step 3: Algebraic Simplification

Expand the factorial terms to identify common factors:

$\frac{5!}{(5-r)!} = 2 \cdot \frac{6 \cdot 5!}{(7-r)(6-r)(5-r)!}$

Cancel $5!$ from both sides and $(5-r)!$ from the denominators:

$1 = \frac{2 \cdot 6}{(7-r)(6-r)}$

$1 = \frac{12}{(7-r)(6-r)}$

Step 4: Solving the Quadratic Equation

Rearrange to form a standard quadratic equation:

$(7-r)(6-r) = 12$

$42 - 7r - 6r + r^2 = 12$

$r^2 - 13r + 42 - 12 = 0$

$r^2 - 13r + 30 = 0$

Factorize the quadratic expression:

$(r - 10)(r - 3) = 0$

Step 5: Verification of Constraints

The roots are $r = 10$ and $r = 3$.

  • If $r = 10$: The term ${}^5P_{10}$ is undefined because $n < r$ ($5 < 10$). [Per the domain constraint $r \leq n$].
  • If $r = 3$: Check feasibility: ${}^5P_3 = \frac{5!}{2!} = 60$. Right side: $2 \cdot {}^6P_{3-1} = 2 \cdot {}^6P_2 = 2 \cdot \frac{6!}{4!} = 2 \cdot (6 \cdot 5) = 60$.

The condition $60 = 60$ holds true.

r value Permutations r=3

Final Solution: r = 3


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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