Problem 81 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve for \(n\). $$_{n+1} P_{3}=4 \cdot_{n} P_{2}$$

Step-by-Step Solution

Answer

Therefore, the solution to the equation is \( n = 3 \).

Step 1: Rewrite the equations using the formula for permutations

In this step, we rewrite the given equations using the formula for permutations. So, \(_{n+1}P_{3} = \frac{(n+1)!}{((n+1)-3)!}\) and \(4 \cdot_{n}P_{2} = 4 \cdot \frac{n!}{(n-2)!}\).

Step 2: Simplify equations

Next, we simplify the above equations. Therefore, \(_{n+1}P_{3} = \frac{(n+1)!}{(n-2)!}\) and \(4 \cdot_{n}P_{2} = 4 \cdot \frac{n!}{(n-2)!}\).

Step 3: Equate both expressions of the permutation

At this stage, we are to equate both expressions of the permutation: \(\frac{(n+1)!}{(n-2)!} = 4 \cdot \frac{n!}{(n-2)!}\).

Step 4: Cancel shared term

Since \((n-2)!\) appears in both sides of the equation, we can cancel it out: \( (n+1)! = 4 \cdot n! \).

Step 5: Express the factorials and simplify

Express the factorials: \( (n+1) \cdot n! = 4 \cdot n! \). This simplifies to \( (n+1) = 4 \).

Step 6: Solve for 'n'

Finally, we solve for \(n\) by subtracting 1 from both sides of the equation: \(n = 4 -1 \)