Question

Find the value of \[{}^6{C_3}\].

Answer 1

Hint: 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 \[0 \leqslant r \leqslant n\].