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CBSE - Class 11 Mathematics Permutations and Combinations Worksheet
EXERCISE 6.2
Worksheet Answers
Solution:
The factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$. The definition is given as: $n! = n \times (n-1) \times (n-2) \times \dots \times 1$ [By the fundamental definition of factorials in combinatorics].
The expression on the left-hand side is $3! + 4!$.
Summing these values: $3! + 4! = 6 + 24 = 30$ [Arithmetic addition].
The expression on the right-hand side is $7!$.
$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
Performing the multiplication sequentially: $7 \times 6 = 42$ $42 \times 5 = 210$ $210 \times 4 = 840$ $840 \times 3 = 2520$ $2520 \times 2 = 5040$ $5040 \times 1 = 5040$ [Arithmetic multiplication].
| Expression | Result |
|---|---|
| $LHS (3! + 4!)$ | $30$ |
| $RHS (7!)$ | $5040$ |
Comparing the calculated values: $30 \neq 5040$.
In general, for any positive integers $m$ and $n$ where $m, n > 1$, $(m! + n!) \neq (m+n)!$ because the factorial function grows super-exponentially. Specifically, by the properties of factorials: $(m+n)! = (m+n) \times (m+n-1) \times \dots \times (n+1) \times n!$ This product is significantly larger than the sum of two individual factorials.
Final Solution: No, $3! + 4! \neq 7!$ because $30 \neq 5040$.
Solution:
The factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$. It is defined by the recurrence relation: $n! = n \times (n-1) \times (n-2) \times \dots \times 1$. [Per the Fundamental Definition of Factorials].
To evaluate the expression $\frac{8!}{6! \times 2!}$, we utilize the property of factorials where $n! = n \times (n-1) \times (n-2)!$. This allows for the cancellation of common terms in the numerator and denominator.
Expanding the numerator until we reach the largest factorial in the denominator:
$8! = 8 \times 7 \times 6!$
Substitute the expanded numerator back into the original expression:
$\text{Expression} = \frac{8 \times 7 \times 6!}{6! \times 2!}$
Cancel the common term $6!$ from the numerator and the denominator: [By the Property of Multiplicative Identity and Division of Like Terms]
$\text{Expression} = \frac{8 \times 7}{2!}$
Evaluate the denominator and perform the final multiplication:
Final Solution: 28
Solution:
Given the equation: $$\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}$$ To simplify the expression, we utilize the property of factorials where $n! = n \times (n-1)!$ [Per the definition of factorial functions]. We express the larger factorials in terms of $6!$:
Substitute the expanded factorials into the equation:
$$\frac{1}{6!} + \frac{1}{7 \times 6!} = \frac{x}{8 \times 7 \times 6!}$$To eliminate the common factor of $\frac{1}{6!}$ from the entire equation, we multiply every term by $6!$ [By the Multiplicative Property of Equality]:
$$1 + \frac{1}{7} = \frac{x}{8 \times 7}$$Evaluate the addition on the LHS by finding a common denominator:
$$\frac{7}{7} + \frac{1}{7} = \frac{x}{56}$$ $$\frac{8}{7} = \frac{x}{56}$$Multiply both sides of the equation by $56$ to solve for $x$ [By the Division/Multiplication Property of Equality]:
$$x = \frac{8}{7} \times 56$$ $$x = 8 \times \left(\frac{56}{7}\right)$$ $$x = 8 \times 8$$ $$x = 64$$Final Solution: The value of $x$ is $64$.
Solution:
The problem requires the evaluation of the factorial expression $\frac{n!}{(n-r)!}$ given the parameters $n = 6$ and $r = 2$. This expression represents the formula for calculating permutations of $n$ items taken $r$ at a time, denoted as $P(n, r) = \frac{n!}{(n-r)!}$. [By the Fundamental Principle of Counting]
Substitute $n = 6$ and $r = 2$ into the expression:
$\frac{6!}{(6-2)!}$
Simplifying the denominator: $6 - 2 = 4$.
The expression becomes: $\frac{6!}{4!}$
Recall the definition of a factorial: $n! = n \times (n-1) \times (n-2) \times \dots \times 1$. [By the definition of the factorial function]
Expand $6!$ until reaching $4!$ to allow for cancellation:
$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$6! = 6 \times 5 \times 4!$
Substitute the expanded form back into the quotient:
$\frac{6 \times 5 \times 4!}{4!}$
Cancel the common factor $4!$ from both the numerator and the denominator: [By the Property of Multiplicative Identity]
$6 \times 5 = 30$
| Component | Value |
|---|---|
| Numerator ($n!$) | $720$ |
| Denominator ($(n-r)!$) | $24$ |
| Result | $30$ |
Final Solution: 30
Solution:
The factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$. Mathematically, it is defined as:
$n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1$
[By the Fundamental Definition of Factorial Notation]
To evaluate the factorial of 4, we perform the following product:
$4! = 4 \times 3 \times 2 \times 1$
$4! = 24$
To evaluate the factorial of 3, we perform the following product:
$3! = 3 \times 2 \times 1$
$3! = 6$
Substituting the computed values of $4!$ and $3!$ into the given expression $4! - 3!$:
Expression $= 24 - 6$
Expression $= 18$
Final Solution: $4! - 3! = 18$
Solution:
The factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers from $1$ up to $n$. Mathematically, it is defined as:
$n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1$
[Per the definition of the factorial function in combinatorial mathematics].
Applying the definition to $n = 8$, we expand the factorial as the product of integers in descending order until reaching unity:
$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
To ensure numerical accuracy, we perform the multiplication in cumulative stages:
By evaluating the product of the first eight natural integers, we arrive at the total number of permutations of eight distinct objects, which is $40,320$.
Final Solution: $8! = 40,320$
Solution:
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 $n$: $n! = n \times (n-1) \times (n-2) \times \dots \times 1$.
Given the variables:
Perform the subtraction within the parentheses: $9 - 5 = 4$ Thus, the expression simplifies to: $$\frac{9!}{4!}$$
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]. $$\text{Expression} = 9 \times 8 \times 7 \times 6 \times 5$$
Executing the multiplication sequentially:
Final Solution: 15120
