Question

Solve the problem: $^6{C_2}$.

Answer 1

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 $^6{C_2}$ as $n = 6$ and $r = 2$. We complete the multiplication and find the solution.

Complete step by step answer:
The given mathematical expression $^6{C_2}$ is an example of a 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 $^6{C_2}$. We put the values of $n = 6$ and $r = 2$ to get,
$^6{C_2} = \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}}$.

Now, we simplify the factorial $6!$ as $6 \times 5 \times 4!$, we get,
${ \Rightarrow ^6}{C_2} = \dfrac{{6!}}{{2! \times 4!}}$
Cancelling out the $4!$ common in both numerator and denominator and putting the value of $2!$, we get,
${ \Rightarrow ^6}{C_2} = \dfrac{{6 \times 5 \times 4!}}{{2! \times 4!}}$
${ \Rightarrow ^6}{C_2} = \dfrac{{6 \times 5}}{{2!}}$
Simplifying the calculations by cancelling common factor in numerator and denominator, we get,
${ \therefore ^6}{C_2} = 15$

Therefore, the value of the combination $^6{C_2}$ is $15$.

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 objects out of n objects. The mathematical expression is $^n{C_{\left( {n - r} \right)}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}{ = ^n}{C_r}$.