Permutation Problem: Determining the Value of n

The problem asks us to find the value of 'n' given a specific equation involving permutations. The equation is presented as a ratio of two permutation expressions: \(\frac{^nP_4}{^{n- 1}P_4} = \frac{5}{3}\), with the condition that \(n > 4\).

Permutation Formula Introduction

To solve this problem, we first need to recall the definition and formula for permutations. A permutation, denoted as \(^nP_r\) (or P(n, r)), represents the number of ways to arrange 'r' distinct items selected from a set of 'n' distinct items, where the order of arrangement matters.

The formula for permutations is:

\[ ^nP_r = \frac{n!}{(n-r)!} \]

where \(n!\) (n factorial) is the product of all positive integers less than or equal to n. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1\).

Applying the Permutation Formula

Let's apply the permutation formula to each term in the given ratio:

• For the numerator, \(^nP_4\), we have \(n\) items and we are choosing \(r=4\) of them.
• Using the formula, \(^nP_4 = \frac{n!}{(n-4)!}\).

• For the denominator, \(^{n-1}P_4\), we have \(n-1\) items and we are choosing \(r=4\) of them.
• Using the formula, \(^{n-1}P_4 = \frac{(n-1)!}{((n-1)-4)!} = \frac{(n-1)!}{(n-5)!}\).

Setting Up the Equation

Now, we substitute these expressions back into the original equation:

\[ \frac{^nP_4}{^{n- 1}P_4} = \frac{\frac{n!}{(n-4)!}}{\frac{(n-1)!}{(n-5)!}} = \frac{5}{3} \]

To simplify the complex fraction, we can multiply the numerator by the reciprocal of the denominator:

\[ \frac{n!}{(n-4)!} \times \frac{(n-5)!}{(n-1)!} = \frac{5}{3} \]

Simplifying Factorial Expressions

To simplify the terms involving factorials, we can expand the larger factorials in terms of smaller ones:

• We know that \(n! = n \times (n-1) \times (n-2) \times \dots \times 1\). Therefore, \(n! = n \times (n-1)!\).
• Similarly, \((n-4)! = (n-4) \times (n-5) \times (n-6) \times \dots \times 1\). Therefore, \((n-4)! = (n-4) \times (n-5)!\).

Substitute these expanded forms into the equation:

\[ \frac{n \times (n-1)!}{(n-4) \times (n-5)!} \times \frac{(n-5)!}{(n-1)!} = \frac{5}{3} \]

Notice that \((n-1)!\) in the numerator and denominator cancel out, and \((n-5)!\) in the numerator and denominator also cancel out.

This leaves us with a much simpler equation:

\[ \frac{n}{n-4} = \frac{5}{3} \]

Solving for n

Now, we need to solve this algebraic equation for 'n'. We can do this by cross-multiplication:

• Multiply the numerator of the left side by the denominator of the right side: \(3 \times n = 3n\).
• Multiply the numerator of the right side by the denominator of the left side: \(5 \times (n-4) = 5(n-4)\).

Setting these two products equal:

\[ 3n = 5(n-4) \]

Distribute the 5 on the right side:

\[ 3n = 5n - 20 \]

Now, gather the terms with 'n' on one side and the constant term on the other side. Subtract \(3n\) from both sides:

\[ 0 = 5n - 3n - 20 \]

\[ 0 = 2n - 20 \]

Add 20 to both sides:

\[ 20 = 2n \]

Finally, divide by 2 to find the value of 'n':

\[ n = \frac{20}{2} \]

\[ n = 10 \]

Verifying the Condition

The problem stated that \(n > 4\). Our calculated value for \(n\) is 10. Since \(10 > 4\), our solution is consistent with the given condition.

Final Answer

Based on our calculations, the value of 'n' that satisfies the given permutation equation is 10.