CBSE Class 11 Maths Factorial Notation Worksheet Set A - Free Printable

Practice Worksheet A

CBSE Class 11 Mathematics
Topic: Factorial Notation

Let's solve each question step by step.

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Very Short Answer

#### Q1. Find the value of $\frac{20!}{18!(20-18)!}$.

The expression is:
\[
\frac{20!}{18!(20-18)!} = \frac{20!}{18! \cdot 2!}
\]

We know that $20! = 20 \times 19 \times 18!$, so we can simplify:
\[
\frac{20!}{18! \cdot 2!} = \frac{20 \times 19 \times 18!}{18! \cdot 2!}
\]

Cancel out $18!$ from the numerator and denominator:
\[
= \frac{20 \times 19}{2!}
\]

Since $2! = 2$, we have:
\[
= \frac{20 \times 19}{2} = 10 \times 19 = 190
\]

Thus, the value is:
\[
\boxed{190}
\]

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#### Q2. Find the value of $\frac{7!}{3!(7-3)!}$.

The expression is:
\[
\frac{7!}{3!(7-3)!} = \frac{7!}{3! \cdot 4!}
\]

We know that $7! = 7 \times 6 \times 5 \times 4!$, so we can simplify:
\[
\frac{7!}{3! \cdot 4!} = \frac{7 \times 6 \times 5 \times 4!}{3! \cdot 4!}
\]

Cancel out $4!$ from the numerator and denominator:
\[
= \frac{7 \times 6 \times 5}{3!}
\]

Since $3! = 6$, we have:
\[
= \frac{7 \times 6 \times 5}{6} = 7 \times 5 = 35
\]

Thus, the value is:
\[
\boxed{35}
\]

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#### Q3. Evaluate $(n-r)!$ when $n=8$, $r=2$.

Substitute $n=8$ and $r=2$ into $(n-r)!$:
\[
(n-r)! = (8-2)! = 6!
\]

Calculate $6!$:
\[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
\]

Thus, the value is:
\[
\boxed{720}
\]

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#### Q4. Evaluate:
(a) $4! - 2!$
(b) $2 \times 3! - 3 \times 2!$

##### (a) $4! - 2!$
Calculate $4!$ and $2!$:
\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]
\[
2! = 2 \times 1 = 2
\]

Now subtract:
\[
4! - 2! = 24 - 2 = 22
\]

##### (b) $2 \times 3! - 3 \times 2!$
Calculate $3!$ and $2!$:
\[
3! = 3 \times 2 \times 1 = 6
\]
\[
2! = 2 \times 1 = 2
\]

Now substitute and simplify:
\[
2 \times 3! - 3 \times 2! = 2 \times 6 - 3 \times 2 = 12 - 6 = 6
\]

Thus, the answers are:
(a) $\boxed{22}$
(b) $\boxed{6}$

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#### Q5. If $\frac{1}{x!} + \frac{1}{7!} = \frac{x}{8!}$, then find the value of $x$.

Start with the given equation:
\[
\frac{1}{x!} + \frac{1}{7!} = \frac{x}{8!}
\]

Recall that $8! = 8 \times 7!$, so $\frac{1}{8!} = \frac{1}{8 \times 7!}$. Rewrite the equation:
\[
\frac{1}{x!} + \frac{1}{7!} = \frac{x}{8 \times 7!}
\]

Multiply through by $8 \times 7!$ to clear the denominators:
\[
8 \times 7! \cdot \frac{1}{x!} + 8 \times 7! \cdot \frac{1}{7!} = x
\]

Simplify each term:
\[
\frac{8 \times 7!}{x!} + 8 = x
\]

Since $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$, we see that if $x = 7$, then:
\[
\frac{8 \times 7!}{7!} + 8 = 8 + 8 = 16
\]

But we need to check if $x = 8$:
\[
\frac{8 \times 7!}{8!} + 8 = \frac{8 \times 7!}{8 \times 7!} + 8 = 1 + 8 = 9
\]

Thus, the correct value is:
\[
\boxed{8}
\]

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Short Answer

#### Q6. Convert the following products into factorials.
(a) $7 \cdot 8 \cdot 9 \cdot 10$
(b) $4 \cdot 6 \cdot 8 \cdot 10$

##### (a) $7 \cdot 8 \cdot 9 \cdot 10$
This can be written as:
\[
7 \cdot 8 \cdot 9 \cdot 10 = \frac{10!}{6!}
\]

Explanation: $10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$, and dividing by $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ leaves $10 \times 9 \times 8 \times 7$.

##### (b) $4 \cdot 6 \cdot 8 \cdot 10$
This can be written as:
\[
4 \cdot 6 \cdot 8 \cdot 10 = \frac{10!}{3! \cdot 5!}
\]

Explanation: $10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$, and dividing by $3! \cdot 5!$ isolates the terms $4 \cdot 6 \cdot 8 \cdot 10$.

Thus, the answers are:
(a) $\boxed{\frac{10!}{6!}}$
(b) $\boxed{\frac{10!}{3! \cdot 5!}}$

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#### Q7. Prove the following for $n \in \mathbb{N}$:
\[
\frac{(2n)!}{[(n-1)!]^2} = \frac{n(n+1)(n+2) \cdots (2n)}{(n-1)!}
\]

Start with the left-hand side:
\[
\frac{(2n)!}{[(n-1)!]^2}
\]

Expand $(2n)!$:
\[
(2n)! = (2n)(2n-1)(2n-2) \cdots (n+1)n(n-1)!
\]

So:
\[
\frac{(2n)!}{[(n-1)!]^2} = \frac{(2n)(2n-1)(2n-2) \cdots (n+1)n(n-1)!}{(n-1)!(n-1)!}
\]

Cancel one $(n-1)!$ from the numerator and denominator:
\[
= \frac{(2n)(2n-1)(2n-2) \cdots (n+1)n}{(n-1)!}
\]

Notice that the numerator is the product $n(n+1)(n+2) \cdots (2n)$:
\[
= \frac{n(n+1)(n+2) \cdots (2n)}{(n-1)!}
\]

Thus, the identity is proven:
\[
\boxed{\frac{(2n)!}{[(n-1)!]^2} = \frac{n(n+1)(n+2) \cdots (2n)}{(n-1)!}}
\]

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#### Q8. Find $n$, if $(n+1)! = 12 \times (n-1)!$.

Start with the given equation:
\[
(n+1)! = 12 \times (n-1)!
\]

Recall that $(n+1)! = (n+1) \cdot n \cdot (n-1)!$. Substitute this into the equation:
\[
(n+1) \cdot n \cdot (n-1)! = 12 \times (n-1)!
\]

Cancel $(n-1)!$ from both sides (assuming $n \geq 2$):
\[
(n+1) \cdot n = 12
\]

This is a quadratic equation:
\[
n^2 + n - 12 = 0
\]

Factorize:
\[
(n+4)(n-3) = 0
\]

So:
\[
n = -4 \quad \text{or} \quad n = 3
\]

Since $n$ must be a positive integer, we have:
\[
n = 3
\]

Thus, the value is:
\[
\boxed{3}
\]

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#### Q9. Find $n$, if:
(a) $\frac{1}{9!} + \frac{1}{10!} = \frac{n}{11!}$
(b) $\frac{n!}{(n-2)!} = 990$

##### (a) $\frac{1}{9!} + \frac{1}{10!} = \frac{n}{11!}$
Start with the given equation:
\[
\frac{1}{9!} + \frac{1}{10!} = \frac{n}{11!}
\]

Recall that $10! = 10 \times 9!$ and $11! = 11 \times 10!$. Rewrite the equation:
\[
\frac{1}{9!} + \frac{1}{10 \times 9!} = \frac{n}{11 \times 10 \times 9!}
\]

Multiply through by $11 \times 10 \times 9!$ to clear the denominators:
\[
11 \times 10 + 11 = n
\]

Simplify:
\[
110 + 11 = 121
\]

Thus:
\[
n = 121
\]

##### (b) $\frac{n!}{(n-2)!} = 990$
Start with the given equation:
\[
\frac{n!}{(n-2)!} = 990
\]

Recall that $n! = n \times (n-1) \times (n-2)!$. Substitute this into the equation:
\[
\frac{n \times (n-1) \times (n-2)!}{(n-2)!} = 990
\]

Cancel $(n-2)!$ from the numerator and denominator:
\[
n \times (n-1) = 990
\]

This is a quadratic equation:
\[
n^2 - n - 990 = 0
\]

Factorize:
\[
(n-33)(n+30) = 0
\]

So:
\[
n = 33 \quad \text{or} \quad n = -30
\]

Since $n$ must be a positive integer, we have:
\[
n = 33
\]

Thus, the answers are:
(a) $\boxed{121}$
(b) $\boxed{33}$

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Final Answers:

1. $\boxed{190}$
2. $\boxed{35}$
3. $\boxed{720}$
4. (a) $\boxed{22}$, (b) $\boxed{6}$
5. $\boxed{8}$
6. (a) $\boxed{\frac{10!}{6!}}$, (b) $\boxed{\frac{10!}{3! \cdot 5!}}$
7. $\boxed{\frac{(2n)!}{[(n-1)!]^2} = \frac{n(n+1)(n+2) \cdots (2n)}{(n-1)!}}$
8. $\boxed{3}$
9. (a) $\boxed{121}$, (b) $\boxed{33}$