Question

How do you evaluate ${}^{12}{C_8}$?

Answer 1

In mathematics, a combination is a selection of items from a collection such that the order of selection does not matter is called combination. In the question we are asked to evaluate the given combination that we will do simply by using the formula for evaluating the combination.

Formula used:
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Where; $r < n$
${}^n{C_r} =$ number of combinations
$n =$ total number of objects in the set
$r =$ number of choosing objects from the set

Complete step by step answer:
We have to evaluate ${}^{12}{C_8}$.
We know that;
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
So, on comparing with ${}^n{C_r}$, we get;
$n = 12$
$r = 8$
Putting these values in the formula we get;
$\Rightarrow {}^{12}{C_8} = \dfrac{{12!}}{{8!\left( {12 - 8} \right)!}}$
Solving the bracket, we get;
$\Rightarrow {}^{12}{C_8} = \dfrac{{12!}}{{8!4!}}$
Simplifying the numerator, we get;
$\Rightarrow {}^{12}{C_8} = \dfrac{{12 \times 11 \times 10 \times 9 \times 8!}}{{8! \times 4!}}$
Cancelling the terms from the numerator and expanding the factorial we get;
$\Rightarrow {}^{12}{C_8} = \dfrac{{12 \times 11 \times 10 \times 9}}{{4 \times 3 \times 2 \times 1}}$
On calculating we get;
$\Rightarrow {}^{12}{C_8} = 495$

Note: When the order of selection does not matter. Then we use combinations and when the order of selection matters then we use the permutation. For example, A, B in combination is the same as B, A but is different in permutation. Also, in the formula we used above to solve the question if $r > n$, then the total combination will be equal to zero. Practically, to find the permutation of $n$ different objects taken $r$ at a time, we first need to select $r$ items from $n$ items and then arrange them. So, usually the number of permutations exceeds the number of combinations.