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# The value of ${}^2{P_1} + {}^3{P_1} + ... + {}^n{P_1}$ is equal toA.$\dfrac{{{n^2} - n + 2}}{2}$B.$\dfrac{{{n^2} + n + 2}}{2}$C.$\dfrac{{{n^2} + n - 1}}{2}$D.$\dfrac{{{n^2} - n - 1}}{2}$E.$\dfrac{{{n^2} + n - 2}}{2}$

Last updated date: 12th Sep 2024
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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 $1 + 2 + 3 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2}$.

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.