Question

If \[5P\left( {4,n} \right) = 6P\left( {5,n - 1} \right)\], find the value of \[n\].

Answer 1

Hint: Here, we need to find the value of \[n\]. We will use the formula of permutations to simplify the given equation. Then, we will simplify the equation to form a quadratic equation. Finally, we will solve this quadratic equation to find the required value of \[n\].

Formula Used:
The number of permutations in which a set of \[n\] objects can be arranged in \[r\] places is given by \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], where no object is repeated. Here, \[{}^n{P_r}\] can be written as \[P\left( {n,r} \right)\].

Complete step-by-step answer:
We will use the formula for permutations to simplify the given equation.
Substituting \[n\] as 4 and \[r\] as \[n\] in the formula for permutations \[{}^n{P_r} = P\left( {n,r} \right) = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], we get
\[ \Rightarrow {}^4{P_n} = P\left( {4,n} \right) = \dfrac{{4!}}{{\left( {4 - n} \right)!}}\]
Substituting \[n\] as 5 and \[r\] as \[n - 1\] in the formula for permutations \[{}^n{P_r} = P\left( {n,r} \right) = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], we get
\[ \Rightarrow {}^5{P_{n - 1}} = P\left( {5,n - 1} \right) = \dfrac{{5!}}{{\left( {5 - \left( {n - 1} \right)} \right)!}}\]
Rewriting the equation, we get
\[ \Rightarrow P\left( {5,n - 1} \right) = \dfrac{{5!}}{{\left( {5 - 1\left( {n - 1} \right)} \right)!}}\]
Multiplying the terms in the denominator using the distributive law of multiplication, we get
\[ \Rightarrow P\left( {5,n - 1} \right) = \dfrac{{5!}}{{\left( {5 - n + 1} \right)!}}\]
Adding the terms in the denominator, we get
\[ \Rightarrow P\left( {5,n - 1} \right) = \dfrac{{5!}}{{\left( {6 - n} \right)!}}\]
Now, we can simplify the given equation.
Substituting \[P\left( {4,n} \right) = \dfrac{{4!}}{{\left( {4 - n} \right)!}}\] and \[P\left( {5,n - 1} \right) = \dfrac{{5!}}{{\left( {6 - n} \right)!}}\] in the equation \[5P\left( {4,n} \right) = 6P\left( {5,n - 1} \right)\], we get
\[ \Rightarrow 5 \times \dfrac{{4!}}{{\left( {4 - n} \right)!}} = 6 \times \dfrac{{5!}}{{\left( {6 - n} \right)!}}\]
The expression of the form \[n!\] can be written as \[n\left( {n - 1} \right)!\] or \[n\left( {n - 1} \right)\left( {n - 2} \right)!\], etc.
Therefore, we can rewrite \[\left( {6 - n} \right)!\] as \[\left( {6 - n} \right)\left( {6 - n - 1} \right)\left( {6 - n - 2} \right)!\].
Thus, we get
\[\left( {6 - n} \right)! = \left( {6 - n} \right)\left( {5 - n} \right)\left( {4 - n} \right)!\]
Substituting \[\left( {6 - n} \right)! = \left( {6 - n} \right)\left( {5 - n} \right)\left( {4 - n} \right)!\] in the equation \[5 \times \dfrac{{4!}}{{\left( {4 - n} \right)!}} = 6 \times \dfrac{{5!}}{{\left( {6 - n} \right)!}}\], we get
\[ \Rightarrow 5 \times \dfrac{{4!}}{{\left( {4 - n} \right)!}} = 6 \times \dfrac{{5!}}{{\left( {6 - n} \right)\left( {5 - n} \right)\left( {4 - n} \right)!}}\]
Multiplying both sides by \[\left( {4 - n} \right)!\], we get
\[ \Rightarrow 5 \times 4! = 6 \times \dfrac{{5!}}{{\left( {6 - n} \right)\left( {5 - n} \right)}}\]
The value of 4! Is 24, and the value of 5! Is 120.
Therefore, we get
\[ \Rightarrow 5 \times 24 = 6 \times \dfrac{{120}}{{\left( {6 - n} \right)\left( {5 - n} \right)}}\]
Multiplying the terms of the expression, we get
\[ \Rightarrow 120 = \dfrac{{720}}{{\left( {6 - n} \right)\left( {5 - n} \right)}}\]
Dividing both sides by 120, we get
\[ \Rightarrow 1 = \dfrac{6}{{\left( {6 - n} \right)\left( {5 - n} \right)}}\]
Rewriting the equation by cross-multiplying, we get
\[ \Rightarrow \left( {6 - n} \right)\left( {5 - n} \right) = 6\]
Multiplying the terms in the parentheses using the distributive law of multiplication, we get
\[ \Rightarrow 30 - 6n - 5n + {n^2} = 6\]
Subtracting 6 from both sides , we get
\[ \Rightarrow 30 - 6n - 5n + {n^2} - 6 = 0\]
Adding and subtracting the like terms in the equation, we get
\[ \Rightarrow 24 - 11n + {n^2} = 0\]
Rewriting the equation, we get
\[ \Rightarrow {n^2} - 11n + 24 = 0\]
Here, we get a quadratic equation. We will split the middle term and factorise to solve the quadratic equation.
Factoring the quadratic equation, we get
\[\begin{array}{l} \Rightarrow {n^2} - 8n - 3n + 24 = 0\\ \Rightarrow n\left( {n - 8} \right) - 3\left( {n - 8} \right) = 0\\ \Rightarrow \left( {n - 3} \right)\left( {n - 8} \right) = 0\end{array}\]
Therefore, either \[n - 3 = 0\] or \[n - 8 = 0\].
Simplifying the expressions, we get
\[n = 3\] or \[n = 8\].
Now, the number of \[n\] objects is always more than \[r\] places in the formula \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\].
If \[n = 8\], then we get \[P\left( {4,8} \right) = {}^4{P_8}\], which is not possible.
Thus, we get
\[ \Rightarrow n = 3\]
Therefore, the value of \[n\] is 3.

Note: We used the term “quadratic equation” in our solution. A quadratic equation is an equation having the highest degree of variable 2. It is of the form \[a{x^2} + bx + c = 0\], where \[a\] is not equal to 0. A quadratic equation has 2 solutions.
We can verify our answer by substituting \[n = 3\] in the given equation.
Substituting \[n = 3\] in the left hand side of the given equation, we get
\[L.H.S. = 5P\left( {4,3} \right)\]
Using the formula for permutations, we get
\[\begin{array}{l} \Rightarrow L.H.S. = 5 \times \dfrac{{4!}}{{\left( {4 - 3} \right)!}}\\ \Rightarrow L.H.S. = 5 \times \dfrac{{24}}{{\left( 1 \right)!}}\end{array}\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow L.H.S. = 5 \times \dfrac{{24}}{1}\\ \Rightarrow L.H.S. = 120\end{array}\]
Substituting \[n = 3\] in the right hand side of the given equation, we get
\[R.H.S. = 6P\left( {5,3 - 1} \right)\]
Subtracting the terms, we get
\[ \Rightarrow R.H.S. = 6P\left( {5,2} \right)\]
Using the formula for permutations, we get
\[\begin{array}{l} \Rightarrow R.H.S. = 6 \times \dfrac{{5!}}{{\left( {5 - 2} \right)!}}\\ \Rightarrow R.H.S. = 6 \times \dfrac{{120}}{{\left( 3 \right)!}}\end{array}\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow R.H.S. = 6 \times \dfrac{{120}}{6}\\ \Rightarrow R.H.S. = 120\end{array}\]
Thus, we can observe that \[L.H.S. = R.H.S.\].
Hence, we have verified our answer.