Question

How do you evaluate $^{8}{{C}_{2}}$?

Answer 1

Hint: In this question we have been given with the combination function which has the general formula as $^{n}{{C}_{r}}$, where $n$ stands for the total number of objects and $r$ is the number of items which has to be selected from the total number of items $n$. In combination the order does not matter. We will use the formula of combination which is $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. We will substitute the values of $n$ and $r$ in the formula and simplify to get the required solution.

Complete step by step answer:
We have the expression given to us as:
${{\Rightarrow }^{8}}{{C}_{2}}$
We can see that the above expression is in the form of combination formula which is $^{n}{{C}_{r}}$ where $n=8$ and $r=2$.
We know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ therefore, on substituting the values of $n$ and $r$ in the formula, we get:
${{\Rightarrow }^{8}}{{C}_{2}}=\dfrac{8!}{2!\left( 8-2 \right)!}$
Now on simplifying the bracket, we get:
${{\Rightarrow }^{8}}{{C}_{2}}=\dfrac{8!}{2!\left( 6 \right)!}$
Now we know that $8!$ can be written as $8\times 7\times 6!$ therefore, on substituting it in the expression, we get:
${{\Rightarrow }^{8}}{{C}_{2}}=\dfrac{8\times 7\times 6!}{2!\left( 6 \right)!}$
On simplifying the expression, we get:
${{\Rightarrow }^{8}}{{C}_{2}}=\dfrac{8\times 7}{2!}$
We know that $2!$ can be written as $2\times 1$ therefore, on substituting, we get:
${{\Rightarrow }^{8}}{{C}_{2}}=\dfrac{8\times 7}{2\times 1}$
On simplifying the expression, we get:
${{\Rightarrow }^{8}}{{C}_{2}}=4\times 7$
On multiplying the terms, we get:
${{\Rightarrow }^{8}}{{C}_{2}}=28$, which is the required solution.

Note: In this question we have used the combination formula. There also exists a permutation formula which is represented as $^{n}{{P}_{r}}$, where $n$ is the total number of items and $r$ is the number of items which are to be selected from the total items. The formula for permutation is $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. The important disjunction between permutation and combination is that the order of the items selected matters in permutation and not in combination.