If x is a positive integer, then = x! & (x+1)! & (x+2)! \\ (x+1)! & (x+2)! & (x+3)! \\ (x+2)! & (x+3)! & (x+4)! is equal to: - Mathematics MCQ Mathematics Mock Test - 9 2026 | ExamTest.live

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If x is a positive integer, then Δ = x! (x + 1)! (x + 2)! (x + 1)! (x + 2)! (x + 3)! (x + 2)! (x + 3)! (x + 4)! is equal to:

Solution & Step-by-step Explanation

Factor out terms from columns:From C_{1} : x!, From C_{2} : (x + 1)!, From C_{3} : (x + 2)! . Δ = (x!) (x + 1)! (x + 2)! 1 x + 1 (x + 2) (x + 1) 1 x + 2 (x + 3) (x + 2) 1 x + 3 (x + 4) (x + 3) .Perform column operations C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C_{2} : Δ = (x!) (x + 1)! (x + 2)! 1 x + 1 (x + 2) (x + 1) 01 2 (x + 2) 01 2 (x + 3) .Expanding along R_{1} : Δ = (x!) (x + 1)! (x + 2)! [1 (2 (x + 3) - 2 (x + 2))] Δ = (x!) (x + 1)! (x + 2)! [2 x + 6 - 2 x - 4] = 2 (x!) (x + 1)! (x + 2)! .

Try it yourself before checking the explanation above.

If x is a positive integer, then Δ = x! (x + 1)! (x + 2)! (x + 1)! (x + 2)! (x + 3)! (x + 2)! (x + 3)! (x + 4)! is equal to:

2 (x!) (x + 1)!

2 (x!) (x + 1)! (x + 2)!

2 (x!) (x + 3)!

None of these

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