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 $${}^n{C}_r={\displaystyle \frac{n!}{r!\left(n-r\right)!}}$$ 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 $${}^n{C}_r={\displaystyle \frac{n!}{r!\left(n-r\right)!}}$$ 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,
$$\Rightarrow {}^n{C}_1={\displaystyle \frac{n!}{1!\left(n-1\right)!}}$$
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,
$$\Rightarrow {}^n{C}_1={\displaystyle \frac{n(n-1)!}{1!\left(n-1\right)!}}$$
Opening $$n!$$ will make it easy to strike out the denominator to further solve the equation. Therefore, we get,
$$\Rightarrow {}^n{C}_1={\displaystyle \frac{n}{1!}}$$
We know that the value of $$1!$$ is 1. Hence by putting it in the equation, we get
$$\Rightarrow {}^n{C}_1={\displaystyle \frac{n}{1}}$$
Therefore, the value of $${}^n{C}_1=n$$

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