How do you evaluate ${}^{11}{{C}_{7}}$ ?

Hint: To evaluate the given question ${}^{11}{{C}_{7}}$ , we will use a formula that is:
$\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ , Where, $0\le r\le n$
And factorial $n$ denoted as $n!$ represents the multiplication of all the numbers up to $n$ as:
$\Rightarrow n!=1\times 2\times 3\times ...\times \left( n-1 \right)\times n$
Similarly,
$\Rightarrow r!=1\times 2\times 3\times ...\times \left( r-1 \right)\times r$

Complete step by step solution:
Since, we have the given question as ${}^{11}{{C}_{7}}$ in the form of ${}^{n}{{C}_{r}}$ . After comparison of ${}^{11}{{C}_{7}}$ and ${}^{n}{{C}_{r}}$, we will the value for $n$ and $r$ as:
$\Rightarrow n=11$
And
$\Rightarrow r=7$
Here, we use the formula for getting the value of given question as:
$\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Now, we will apply the value$11$ and $7$ in the place of $n$ and $r$ respectively as:
\[\Rightarrow {}^{11}{{C}_{7}}=\dfrac{11!}{7!\left( 11-7 \right)!}\]
Since, we replaced the value of $n$ and $r$ in the formula. Then we solve the bracketed numbers by using subtraction. Thus, we will get $4$ after subtracting $7$ from $11$ as:
$\Rightarrow 11-7=4$
So, we will use this value in the formula as:
\[\Rightarrow {}^{11}{{C}_{7}}=\dfrac{11!}{7!\times 4!}\]
Now, we will expand the all the factorial numbers respectively as:
$\Rightarrow 11!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10\times 11$
Here we can write $7!$ for the multiplication of first $7$ numbers that will help to cancel out $7!$ in the formula as:
$\Rightarrow 11!=7!\times 8\times 9\times 10\times 11$
Further we will do multiplication for next $4$ numbers and will get $7920$ after multiplication. So we can write the above expression as:
$\Rightarrow 11!=7!\times 7920$
Similarly, we will do the expression and multiplication for $4!$ as:
$\Rightarrow 4!=1\times 2\times 3\times 4$
$\Rightarrow 4!=24$
Now, we will apply the values of all factorial except $7!$ because we will eliminate $7!$ as:
\[\Rightarrow {}^{11}{{C}_{7}}=\dfrac{7!\times 7920}{7!\times 24}\]
Here, we wrote $11!$ into the form of multiple of $7!$ in the above formula so that we can cancel out $7!$ as:
\[\Rightarrow {}^{11}{{C}_{7}}=\dfrac{7920}{24}\]
Now, we will divide $7920$ from $24$ that will divide the number $7920$ completely and will get quotient $330$ as:
\[\Rightarrow {}^{11}{{C}_{7}}=330\]
Hence, the solution for the ${}^{11}{{C}_{7}}$ is $330$ .

Note: Since, we read the combination with permutation in the chapter named as permutation and combination, we need to completely understand the difference between these two. Permutation is the method that helps us to know the possible numbers of way for arranging the data in a sequence, while combination helps us to know the possible numbers of way of section of data and both have different formula that we need to learn that are
For permutation:
$\Rightarrow {}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
And for combination:
$\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$