Question

The value of $${}^2{P}_1+{}^3{P}_1+...+{}^n{P}_1$$ is equal to
A.$${\displaystyle \frac{n^2-n+2}{2}}$$
B.$${\displaystyle \frac{n^2+n+2}{2}}$$
C.$${\displaystyle \frac{n^2+n-1}{2}}$$
D.$${\displaystyle \frac{n^2-n-1}{2}}$$
E.$${\displaystyle \frac{n^2+n-2}{2}}$$

Answer 1

Hint: Here, we will use formula of permutation, $${}^n{P}_r={\displaystyle \frac{n!}{\left(n-r\right)!}}$$, where $$n$$ is the number of items, and $$r$$ represents the number of items being chosen. Then we will simplify it using $$1+2+3+...+n={\displaystyle \frac{n\left(n+1\right)}{2}}$$.

Complete step-by-step answer:
We are given that $${}^2{P}_1+{}^3{P}_1+...+{}^n{P}_1$$.
We know that the formula of permutation, $${}^n{P}_r={\displaystyle \frac{n!}{\left(n-r\right)!}}$$, where $$n$$ is the number of items, and $$r$$ represents the number of items being chosen.
Using the above formula in the terms of the given series, we get
$$\begin{gathered} \Rightarrow {\displaystyle \frac{2!}{\left(2-1\right)!}}+{\displaystyle \frac{3!}{\left(3-1\right)!}}+...+{\displaystyle \frac{n!}{\left(n-1\right)!}} \\ \Rightarrow {\displaystyle \frac{2!}{1!}}+{\displaystyle \frac{3!}{2!}}+...+{\displaystyle \frac{n!}{\left(n-1\right)!}} \end{gathered}$$
Simplifying the factorials in the above expression, we get
$$\Rightarrow {\displaystyle \frac{2\times 1!}{1!}}+{\displaystyle \frac{3\times 2!}{2!}}+...+{\displaystyle \frac{n\times \left(n-1\right)!}{\left(n-1\right)!}}$$
We will now cancel the same factorials in numerators and denominators in the above expression, we get
$$\Rightarrow 2+3+...+n$$
Adding and subtracting the above equation with 1, we get
$$\Rightarrow 1+2+3+...+n-1$$
Using the formula, $$1+2+3+...+n={\displaystyle \frac{n\left(n+1\right)}{2}}$$ in the above equation and simplifying, we get
$$\begin{gathered} \Rightarrow {\displaystyle \frac{n\left(n+1\right)}{2}}-1 \\ \Rightarrow {\displaystyle \frac{n\left(n+1\right)-2}{2}} \\ \Rightarrow {\displaystyle \frac{n^2+n-2}{2}} \end{gathered}$$
Hence, option E is correct.

Note: In solving these types of questions, you should be familiar with the formula of permutations. Some students use the formula of combinations, $${}^n{C}_r={\displaystyle \frac{\underset{―}{n}|}{\underset{―}{r}|\cdot \underset{―}{n-r}|}}$$ instead of permutations is $${}^n{P}_r={\displaystyle \frac{n!}{\left(n-r\right)!}}$$, where $$n$$ is the number of items, and $$r$$ represents the number of items being chosen, which is wrong.