Find the positive integer value of n such that ^{n}P_4 = 42 \times {^{n}P_2}. — JAMB UTME

Find the positive integer value of $n$ such that $^{n}P_4 = 42 \times {^{n}P_2}$ .

Answer: 9

The positive integer value of

n

is 9.

Expanding

^{n}P_4

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

and

^{n}P_2

n(n-1)

allows dividing out

n(n-1)

, leading to

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

. Expanding and factoring gives

n^2 - 5n - 36 = 0

, which yields

n = 9

as the only valid positive integer.

Apply the permutation formula

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

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

and

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

By definition of permutations, selecting

r

items from

n

distinct items without replacement.

Substitute the expansions into the given relation

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

Direct substitution into

^{n}P_4 = 42 \times {^{n}P_2}

Simplify by dividing out common non-zero terms

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

Since

n \ge 4

n(n-1) \neq 0

and can be safely divided from both sides.

Form and solve the quadratic equation

n^2 - 5n + 6 = 42 \implies n^2 - 5n - 36 = 0 \implies (n-9)(n+4) = 0

Expanding terms and factoring the resulting quadratic expression.

Determine the valid root

n = 9

Permutation total items

n

must satisfy

n \ge r \ge 0

, rejecting the negative root

n = -4

Algebraic equations involving permutations

Estimated Time:1m 15s

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