Question

The value of \[{}^n{c_1}\] is ___________

Answer 1

Hint: The question given above is from permutation and combinations. To balance out the equation, use the combination formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] and insert the value of \[r = 1\] in the formula, then solving it by opening the factorial of the numerator.

Complete step by step answer:
We have a combination formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] which we will use here and r is given to be 1.
So, we will apply the same formula to calculate \[{}^n{c_1}\].
By putting the values in the given formula, we get,
\[ \Rightarrow {}^n{C_1} = \dfrac{{n!}}{{1!\left( {n - 1} \right)!}}\]
So, as you can see there are no common values that we can strike out to balance the whole formula so we will open the factorial of the numerator, which is n.
\[n! = n\left( {n - 1} \right)!\]
By putting this value in the equation, we get,
\[ \Rightarrow {}^n{C_1} = \dfrac{{n(n - 1)!}}{{1!\left( {n - 1} \right)!}}\]
Opening \[n!\] will make it easy to strike out the denominator to further solve the equation. Therefore, we get,
\[ \Rightarrow {}^n{C_1} = \dfrac{n}{{1!}}\]
We know that the value of \[1!\] is 1. Hence by putting it in the equation, we get
\[ \Rightarrow {}^n{C_1} = \dfrac{n}{1}\]
Therefore, the value of \[{}^n{C_1} = {\text{ }}n\]

Note:
Here, the exclamation mark \[\left( ! \right)\] represents factorial. As soon as the value of n and r changes, we will substitute that value into the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]. Also, always try to open the factorials whenever there is nothing to strike out in the equation to simplify it.