Question

Find the value of ${}^6{C_3}$.

Answer 1

The number of all combinations of $n$ distinct objects taken $r$ at a time is given by ${}^n{C_r}$ $=$ $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.

Complete step-by-step answer:
The term $C$ in the expression ${}^6{C_3}$ denotes Combination. By definition, each of the different selections made by taking some or all of a number of objects, irrespective of their arrangement is called combination.
The number of all combinations of $n$ distinct objects taken $r$ at a time is given by
${}^n{C_r}$ $=$ $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
$\therefore {}^6{C_3}$ $=$ $\dfrac{{6!}}{{3!\left( {6 - 3} \right)!}}$ [Comparing with the above formula observe here, $n = 6$, $r = 3$]
$\Rightarrow {}^6{C_3}$ $=$ $\dfrac{{6!}}{{3! \cdot 3!}}$
$\Rightarrow {}^6{C_3}$ $=$ $\dfrac{{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}{{\left( {3 \cdot 2 \cdot 1} \right) \cdot \left( {3 \cdot 2 \cdot 1} \right)}}$
Cancel out the common factors and simplify:
$\Rightarrow {}^6{C_3}$ $=$ $\dfrac{{5 \cdot 4}}{1}$
$\Rightarrow {}^6{C_3}$ $=$ $20$
Hence, the value of ${}^6{C_3}$ is $20$.

Additional information:
${}^n{C_r}$ is also denoted as $C\left( {n,r} \right)$ or \left( {\begin{array}{*{20}{c}} n \\\ r \end{array}} \right), and ${}^n{C_r}$ $=$ $\dfrac{{{}^n{P_r}}}{{r!}}$ , where $P$ denotes permutation and ${}^n{P_r}$ $=$ $\dfrac{{n!}}{{\left( {n - r} \right)!}}$.

Note: The factorial of a number denoted by $n!$ is equal to the product of all numbers from $1$ to that number. For example $4! = 4 \times 3 \times 2 \times 1 = 24$ and $3! = 3 \times 2 \times 1 = 6$.
${}^n{C_r}$ is defined only when $n$ and $r$ are non-negative integers such that $0 \leqslant r \leqslant n$.