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Q5(ii):
Evaluate $frac{n!}{(n-r)!}$, when (ii) $n = 9, r = 5$.

Solution :

Step 1: Definition of Factorial and Permutation Formula

The expression provided is $\frac{n!}{(n-r)!}$, which mathematically represents the number of permutations of $n$ distinct objects taken $r$ at a time, denoted as $P(n, r)$. By definition, the factorial of a non-negative integer $n$ is the product of all positive integers less than or equal to $n$: $n! = n \times (n-1) \times (n-2) \times \dots \times 1$.

Step 2: Substitution of Given Values

Given the variables:

  • $n = 9$
  • $r = 5$
Substituting these values into the expression $\frac{n!}{(n-r)!}$, we obtain: $$\frac{9!}{(9-5)!}$$

Step 3: Simplification of the Denominator

Perform the subtraction within the parentheses: $9 - 5 = 4$ Thus, the expression simplifies to: $$\frac{9!}{4!}$$

Step 4: Algebraic Expansion and Cancellation

Per the property of factorials $n! = n \times (n-1) \times \dots \times (k+1) \times k!$, we expand $9!$ until we reach $4!$ to facilitate the reduction of the fraction: $$\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!}$$ [By the definition of factorial division, common factors in the numerator and denominator can be cancelled provided the denominator is non-zero]. $$\text{Expression} = 9 \times 8 \times 7 \times 6 \times 5$$

Step 5: Final Computation

Executing the multiplication sequentially:

  • $9 \times 8 = 72$
  • $72 \times 7 = 504$
  • $504 \times 6 = 3024$
  • $3024 \times 5 = 15120$

Visualization of Permutation Logic

9 × 8 × 7 × 6 × 5 = 15120 Calculation of P(9, 5):

Final Solution: 15120



CBSE Solutions for Class 11 Mathematics Permutations and Combinations


Chapters in CBSE - Class 11 Mathematics


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