Question

Find \[n\] if \[{}^7{P_3} = n \cdot {}^7{C_3}\]

Answer 1

Hint:
Here we need to find the value of the given variable in the expression. We will use the formula of permutation in the left hand side expression and then we will use the formula of combination in the right hand side expression. From there, we will get the equation including the variable. We will solve the equation to get the value of the required variable.

Complete Step by Step Solution:
The given expression is \[{}^7{P_3} = n \cdot {}^7{C_3}\]
Now, we will use the formula of permutation, \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] in the left hand side expression. Therefore, we get
\[ \Rightarrow \dfrac{{7!}}{{\left( {7 - 3} \right)!}} = n \cdot {}^7{C_3}\]
Now, using the formula of combination \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] in the left hand side expression, we get
\[ \Rightarrow \dfrac{{7!}}{{\left( {7 - 3} \right)!}} = n \cdot \dfrac{{7!}}{{3!\left( {7 - 3} \right)!}}\]
Subtracting the terms in the denominator, we get
\[ \Rightarrow \dfrac{{7!}}{{4!}} = n \cdot \dfrac{{7!}}{{3! \times 4!}}\]
Dividing both sides by \[7!\], we get
\[ \Rightarrow \dfrac{1}{{4!}} = n \cdot \dfrac{1}{{3! \times 4!}}\]
Now, multiplying \[4!\] on both sides, we get
\[\begin{array}{l} \Rightarrow \dfrac{1}{{4!}} \times 4! = n \cdot \dfrac{1}{{3! \times 4!}} \times 4!\\ \Rightarrow 1 = \dfrac{n}{{3!}}\end{array}\]
On cross multiplying the terms, we get
\[ \Rightarrow 3! = n\]
Computing the factorial of 3, we get
\[ \Rightarrow 3 \times 2 \times 1 = n\]
On multiplying the terms, we get
\[ \Rightarrow 6 = n\]

Therefore, the value of \[n\] is equal to 6.

Additional information:
Factorial of any positive integer can be defined as the multiplication of all the positive integers less than or equal to the given positive integers. Factorial of zero is one. Factorials are generally used in permutations and combinations problems. Factorials of negative integers are not defined.

Note:
There is an alternate solution to solve this problem. Here we will use the relation between the formula of combination and the permutation.
The given expression is \[{}^7{P_3} = n \cdot {}^7{C_3}\].
We know the formula;
\[{}^n{C_r} \times r! = {}^n{P_r}\]
Now, we will use this formula in the left hand side expression.
\[ \Rightarrow {}^7{C_3} \times 3! = n \cdot {}^7{C_3}\]
On further simplification, we get
\[ \Rightarrow 3! = n\]
 Now, we will put the value of the factorial of 3.
\[ \Rightarrow 3 \times 2 \times 1 = n\]
On multiplying the terms, we get
\[ \Rightarrow 6 = n\]
Therefore, the value of \[n\] is equal to 6.