Question

The value of \[{}^2{P_1} + {}^3{P_1} + ... + {}^n{P_1}\] is equal to
A.\[\dfrac{{{n^2} - n + 2}}{2}\]
B.\[\dfrac{{{n^2} + n + 2}}{2}\]
C.\[\dfrac{{{n^2} + n - 1}}{2}\]
D.\[\dfrac{{{n^2} - n - 1}}{2}\]
E.\[\dfrac{{{n^2} + n - 2}}{2}\]

Answer 1

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

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, \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], 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
\[
   \Rightarrow \dfrac{{2!}}{{\left( {2 - 1} \right)!}} + \dfrac{{3!}}{{\left( {3 - 1} \right)!}} + ... + \dfrac{{n!}}{{\left( {n - 1} \right)!}} \\
   \Rightarrow \dfrac{{2!}}{{1!}} + \dfrac{{3!}}{{2!}} + ... + \dfrac{{n!}}{{\left( {n - 1} \right)!}} \\
 \]
Simplifying the factorials in the above expression, we get
\[ \Rightarrow \dfrac{{2 \times 1!}}{{1!}} + \dfrac{{3 \times 2!}}{{2!}} + ... + \dfrac{{n \times \left( {n - 1} \right)!}}{{\left( {n - 1} \right)!}}\]
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, \[1 + 2 + 3 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2}\] in the above equation and simplifying, we get
\[
   \Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2} - 1 \\
   \Rightarrow \dfrac{{n\left( {n + 1} \right) - 2}}{2} \\
   \Rightarrow \dfrac{{{n^2} + n - 2}}{2} \\
 \]
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, \[{}^n{C_r} = \dfrac{{\left. {\underline {\,
 n \,}}\! \right| }}{{\left. {\underline {\,
 r \,}}\! \right| \cdot \left. {\underline {\,
 {n - r} \,}}\! \right| }}\] instead of permutations is \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], where \[n\] is the number of items, and \[r\] represents the number of items being chosen, which is wrong.