Given the permutation equation {^{n}P_4} = 42 \cdot {^{n}P_2}, what is the positive intege… — JAMB UTME

Given the permutation equation ${^{n}P_4} = 42 \cdot {^{n}P_2}$ , what is the positive integer value of $n$ ?

Answer: 9

The positive integer value of

n

9

Expanding

{^{n}P_4}

n(n-1)(n-2)(n-3)

and

{^{n}P_2}

n(n-1)

transforms the equation to

n(n-1)(n-2)(n-3) = 42n(n-1)

. Dividing both sides by

n(n-1)

gives

(n-2)(n-3) = 42

, which simplifies to

n^2 - 5n - 36 = 0

. Factoring this quadratic gives

(n-9)(n+4) = 0

. Since

n

must be a positive integer greater than or equal to

4

, the correct answer is

9

Write out the expanded expressions for

{^{n}P_4}

and

{^{n}P_2}

{^{n}P_4} = n(n-1)(n-2)(n-3)

and

{^{n}P_2} = n(n-1)

By definition,

{^{n}P_r} = \frac{n!}{(n-r)!} = n(n-1)\dots(n-r+1)

Set up the algebraic equation based on the given problem statement.

n(n-1)(n-2)(n-3) = 42n(n-1)

Substitute the expanded permutation formulas into the given identity.

Simplify the equation by dividing both sides by the non-zero common product

n(n-1)

(n-2)(n-3) = 42

Since

n \ge 4

n(n-1) \neq 0

, so we can cancel these terms from both sides.

Expand and rearrange the resulting expression into a standard quadratic equation.

n^2 - 5n - 36 = 0

Expanding

(n-2)(n-3)

gives

n^2 - 5n + 6

; subtracting

42

yields

n^2 - 5n - 36 = 0

Solve the quadratic equation for

n

and discard non-physical roots.

n = 9

Factoring yields

(n-9)(n+4) = 0

, so

n = 9

n = -4

. A permutation requires

n

to be a positive integer

\ge 4

, so

n = 9

Solving algebraic equations involving permutations

{^{n}P_r} = \frac{n!}{(n-r)!}

Question