Question

The value of ${}^6{C_4}$ is-
A) $6$
B) $9$
C) $15$
D) $240$

Answer 1

Hint:
The number of ways to select r things from n things is represented as ${}^n{C_r}$ and also we know that the formula of combination is-
${}^n{C_r} = \dfrac{{n!}}{{r!n - r!}}$ where n= total number of things and r is the number of things to be selected and also $n! = n\left( {n - 1} \right)...3, 2, 1$. Use this formula to solve the given question.

Complete step by step solution:
We have to find the value of ${}^6{C_4}$.
We know that the formula of combination is-
$ \Rightarrow {}^n{C_r} = \dfrac{{n!}}{{r!n - r!}}$-- (i)
Where n= total number of things and r is the number of things to be selected.
So here n=$6$and r=$4$
So on putting these values in eq. (i), we get-
 $ \Rightarrow {}^6{C_4} = \dfrac{{6!}}{{4!6 - 4!}}$
On performing subtraction in the denominator, we get-
$ \Rightarrow {}^6{C_4} = \dfrac{{6!}}{{4!2!}}$
Now we know that $n! = n\left( {n - 1} \right)...3,2,1$
So we can write-
$ \Rightarrow {}^6{C_4} = \dfrac{{6 \times \left( {6 - 1} \right) \times \left( {6 - 2} \right) \times \left( {6 - 3} \right) \times \left( {6 - 4} \right) \times \left( {6 - 5} \right)}}{{4 \times \left( {4 - 1} \right) \times \left( {4 - 2} \right) \times \left( {4 - 3} \right) \times 2 \times \left( {2 - 1} \right)}}$
On solving, we get-
$ \Rightarrow {}^6{C_4} = \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1 \times 2 \times 1}}$
On cancelling the same terms of numerator and denominator, we get-
$ \Rightarrow {}^6{C_4} = \dfrac{{6 \times 5}}{{2 \times 1}}$
On multiplication, we get-
$ \Rightarrow {}^6{C_4} = \dfrac{{30}}{2}$
On dividing the numerator by denominator, we get-
$ \Rightarrow {}^6{C_4} = 15$

Hence the correct answer is option C.

Note:
Students may get confused between the formula of combination and permutation as both look almost the same but there is a difference between the two formulae. The combination is only concerned with selection not order while in permutation order is important. A permutation is concerned with the arrangement of things and it is given as-
$ \Rightarrow {}^n{P_r} = \dfrac{{n!}}{{n - r!}}$
Where n is the total number of things and r is the number of things to be selected. So we can also write the formula of combination as-
$ \Rightarrow {}^n{C_r} = \dfrac{{{}^n{P_r}}}{{r!}}$
So we can also use this formula to solve the given question.