Question

Compute the given factorials:
(i) $\dfrac{10!}{3!7!}$
(ii) $\dfrac{8!}{\left( 6-4 \right)!}$

Answer 1

For problem (i) we need to calculate the values of the given expressions. We can observe that the given expression has the special type of function which is factorial function. We know that the factorial value of integer $n$ can be written as $n!=n\left( n-1 \right)\left( n-2 \right).....3\times 2\times 1$. So, we will list all the values which are having the factorial function and calculate those values by using the above formula. After that we will apply the given arithmetic operations and simplify the expressions to get the required values.
For the problem (ii) we will simplify the denominator by applying the proper arithmetic operation and then calculate the factorial value of each term which are in numerator and denominator individually and then we will simplify the expression to get the required result.

Complete step-by-step solution:
In (i) we have given the expression $\dfrac{10!}{3!7!}$.
In the above expression we can observe the values $10!$, $7!$, $3!$.
We know that the factorial value of the integer $n$ will be $n!=n\left( n-1 \right)\left( n-2 \right).....3\times 2\times 1$.
From the above formula we can write the value of $10!$ as
$\begin{aligned} & \Rightarrow 10!=10\times \left( 10-1 \right)\times \left( 10-2 \right)\times \left( 10-3 \right)\times \left( 10-4 \right)\times ...3\times 2\times 1 \\\ & \Rightarrow 10!=10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \\\ & \Rightarrow 10!=3628800 \\\ \end{aligned}$
Now the value of $7!$ will be
$\begin{aligned} & \Rightarrow 7!=7\times \left( 7-1 \right)\times \left( 7-2 \right)\times \left( 7-3 \right)\times \left( 7-4 \right)\times \left( 7-5 \right)\times \left( 7-6 \right) \\\ & \Rightarrow 7!=7\times 6\times 5\times 4\times 3\times 2\times 1 \\\ & \Rightarrow 7!=5040 \\\ \end{aligned}$
And the value of $3!$ will be
$\begin{aligned} & \Rightarrow 3!=3\times \left( 3-1 \right)\times \left( 3-2 \right) \\\ & \Rightarrow 3!=3\times 2\times 1 \\\ & \Rightarrow 3!=6 \\\ \end{aligned}$
Substituting the all the values we have calculated, in the given expression, then we will get
$\begin{aligned} & \Rightarrow \dfrac{10!}{3!7!}=\dfrac{3628800}{6\times 5040} \\\ & \Rightarrow \dfrac{10!}{3!7!}=120 \\\ \end{aligned}$
Hence the value of the given expression $\dfrac{10!}{3!7!}$ is $120$.
In the problem (ii) we have given the expression $\dfrac{8!}{\left( 6-4 \right)!}$.
Simplifying the denominator which is $\left( 6-4 \right)!$, then we will get
$\Rightarrow \dfrac{8!}{\left( 6-4 \right)!}=\dfrac{8!}{2!}$
In the above expression we have the values $8!$, $2!$.
We know that the factorial value of the integer $n$ will be $n!=n\left( n-1 \right)\left( n-2 \right).....3\times 2\times 1$.
From the above formula we can write the value of $8!$ as
$\begin{aligned} & \Rightarrow 8!=8\times \left( 8-1 \right)\times \left( 8-2 \right)\times \left( 8-3 \right)\times \left( 8-4 \right)\times ...3\times 2\times 1 \\\ & \Rightarrow 8!=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \\\ & \Rightarrow 8!=40320 \\\ \end{aligned}$
And the value of $2!$ will be
$\begin{aligned} & \Rightarrow 2!=2\times \left( 2-1 \right) \\\ & \Rightarrow 2!=2\times 1 \\\ & \Rightarrow 2!=2 \\\ \end{aligned}$
Substituting the all the values we have calculated, in the given expression, then we will get
$\begin{aligned} & \Rightarrow \dfrac{8!}{\left( 6-4 \right)!}=\dfrac{8!}{2!} \\\ & \Rightarrow \dfrac{8!}{\left( 6-4 \right)!}=\dfrac{40320}{2} \\\ & \Rightarrow \dfrac{8!}{\left( 6-4 \right)!}=20160 \\\ \end{aligned}$
Hence the value of the given expression $\dfrac{8!}{\left( 6-4 \right)!}$ is $20160$.

Note: For (i) We can also solve this problem in another method i.e., we can write the value $10!$ as bellow
$\begin{aligned} & \Rightarrow 10!=10\times \left( 10-1 \right)\times \left( 10-2 \right)\times \left( 10-3 \right)\times \left( 10-4 \right)\times ...3\times 2\times 1 \\\ & \Rightarrow 10!=10\times 9\times 8\times \left( 7\times 6\times 5\times 4\times 3\times 2\times 1 \right) \\\ \end{aligned}$
We can write the value $7\times 6\times 5\times 4\times 3\times 2\times 1$ as $7!$ in the above equation, then we will get
$\Rightarrow 10!=720\times 7!$
Divide the above equation with $7!$ on both sides, then we will get
$\Rightarrow \dfrac{10!}{7!}=720$
Again, divide the above equation with $3!=3\times 2\times 1=6$ on both sides of the above equation, then we will have
$\begin{aligned} & \Rightarrow \dfrac{10!}{3!7!}=\dfrac{720}{6} \\\ & \Rightarrow \dfrac{10!}{3!7!}=120 \\\ \end{aligned}$
From both the methods we got the same result.
For (ii) also we can follow the above method. That is we can write the value $8!$ as bellow
$\begin{aligned} & \Rightarrow 8!=8\times \left( 8-1 \right)\times \left( 8-2 \right)\times \left( 8-3 \right)\times \left( 8-4 \right)\times ...3\times 2\times 1 \\\ & \Rightarrow 8!=8\times 7\times 6\times 5\times 4\times 3\times \left( 2\times 1 \right) \\\ & \Rightarrow 8!=8\times 7\times 6\times 5\times 4\times 3\times 2! \\\ \end{aligned}$
Divide the above equation with $2!$ on both sides, then we will get
$\Rightarrow \dfrac{8!}{2!}=20160$
From both the methods we got the same result.