Question

What is the value of \[{}^{4}{{C}_{2}}\]?

Answer 1

Hint: In the question we have been asked to find the combination of 4 and 2. For solving the problems of combinations we use the formula that is given as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\cdot \left( n-r \right)!}$. This formula is used to calculate combination between 2 numbers.

Complete step-by-step answer:
We have the question with us that wants us to calculate the value of \[{}^{4}{{C}_{2}}\].
To find the value of the required expression we should use the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\cdot \left( n-r \right)!}$. We should substitute 4 in the place of n and substitute 2 in the place of r.
$\begin{align}
  & \Rightarrow {}^{4}{{C}_{2}}=\dfrac{4!}{2!\cdot \left( 4-2 \right)!} \\
 & \Rightarrow {}^{4}{{C}_{2}}=\dfrac{4!}{2!\cdot 2!}......\left( 1 \right) \\
\end{align}$
Now we have obtained the expression (1). In this expression find the value of the factorial of the numbers 4 and 2 or it could be written as $4!$ and$2!$. We know how to find the factorial of a number, it is the product of all the numbers between 1 and the given number both inclusive.
\[\begin{align}
  & 4!=4\cdot 3\cdot 2\cdot 1=24 \\
 & 2!=2\cdot 1=2 \\
\end{align}\]
Now, we have to put the values of the factorial of 4 and 2 in the expression (1). Doing so, we get:
$\begin{align}
  & \Rightarrow {}^{4}{{C}_{2}}=\dfrac{4!}{2!\cdot 2!} \\
 & \Rightarrow {}^{4}{{C}_{2}}=\dfrac{24}{2\cdot 2} \\
 & \Rightarrow {}^{4}{{C}_{2}}=\dfrac{24}{4}....\left( 2 \right) \\
\end{align}$
Now we have obtained the expression (2). In this expression we have to divide 24 by 4. If we divide 24 by 4 then we get 6.
$\Rightarrow {}^{4}{{C}_{2}}=6$
Hence we have obtained the value of the combination that is 6 or ${}^{4}{{C}_{2}}=6$.

Note: We should never get confused with finding the permutation and finding the combination. These are two different things but have similar formulas. These both are used in calculating probability. We also use combinations in binomial theorem.