Question

How do you find the value of binomial coefficient of $ ^8{C_6} $ ?

Answer 1

Hint: We first discuss the general form of combination and its general meaning with the help of variables. We express the mathematical notion with respect to the factorial form of $ ^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ . Then, we place the values for $ ^8{C_6} $ as $ n = 8 $ and $ r = 6 $ . We complete the multiplication and find the solution.

Complete step by step solution:
The given mathematical expression $ ^8{C_6} $ is an example of combination.
We first try to find the general form of combination and its general meaning and then we put the values to find the solution.
The general form of combination is $ ^n{C_r} $ . It’s used to express the notion of choosing r objects out of n objects. The value of $ ^n{C_r} $ expresses the number of ways the combination of those objects can be done.
The simplified form of the mathematical expression $ ^n{C_r} $ is $ ^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ .
Here the term $ n! $ defines the notion of multiplication of first n natural numbers.
This means $ n! = 1 \times 2 \times 3 \times ....n $ .
The arrangement of those chosen objects is not considered in case of combination. That part is involved in permutation.
Now we try to find the value of $ ^8{C_6} $ . We put the values of $ n = 8 $ and $ r = 6 $ to get
\[{ \Rightarrow ^8}{C_6} = \dfrac{{8!}}{{6!\left( {8 - 6} \right)!}}\].
We now solve the factorial values.
\[{ \Rightarrow ^8}{C_6} = \dfrac{{8 \times 7 \times 6!\,}}{{6!\, \times 2!}}\]
Cancelling out the $ 6! $ common in both numerator and denominator and putting the value of $ 2! $ , we get,
\[{ \Rightarrow ^8}{C_6} = \dfrac{{8 \times 7\,}}{2}\]
\[{ \Rightarrow ^8}{C_6} = 4 \times 7\]
Simplifying the expression further, we get,
\[{ \Rightarrow ^8}{C_6} = 28\]
Therefore, the value of the combination $ ^8{C_6} $ is $ 28 $ .
So, the correct answer is “28”.

Note: There are some constraints in the form of $ ^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ . Also, we need to remember the fact that the notion of choosing r objects out of n objects is exactly equal to the notion of choosing $ \left( {n - r} \right) $ objects out of n objects. The mathematical expression is $ ^n{C_{\left( {n - r} \right)}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}{ = ^n}{C_r} $ .