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19 Aug 2026, 22:18
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Which of the following is equal to \(n!*(n+1)!\)?
A. \((n!)^2\)
B. \((n+1)!^2\)
C. \(n!^3(n+1)!^2\)
D. \(n!^3(n+1)\)
E. \(n!^2(n+1)\)
A. \((n!)^2\)
B. \((n+1)!^2\)
C. \(n!^3(n+1)!^2\)
D. \(n!^3(n+1)\)
E. \(n!^2(n+1)\)
20 Aug 2026, 02:29
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Ans - Option E
i took the help of numbers
took two similar number for the variable n, 8 and 9
where n! = 8! and n+1! = 9! so
8!*9! = 8!*(8!*9) [ 9! can be expanded as 9*8!]
so 8!^2(8+1) [ 9 which is 8+1]
so replacing with variable n
n!^2(n+1)
i took the help of numbers
took two similar number for the variable n, 8 and 9
where n! = 8! and n+1! = 9! so
8!*9! = 8!*(8!*9) [ 9! can be expanded as 9*8!]
so 8!^2(8+1) [ 9 which is 8+1]
so replacing with variable n
n!^2(n+1)
Bunuel
muskang2000
Joined: 22 Nov 2024
Last visit: 03 Sep 2026
Posts: 36
Given Kudos: 7
Location: India
Schools: ISB HEC Sept '27 IIM Anderson '27
GMAT Focus 1: 605 Q85 V79 DI76 
GMAT 1: 620 Q44 V31
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Schools: ISB HEC Sept '27 IIM Anderson '27
GMAT Focus 1: 605 Q85 V79 DI76
GMAT 1: 620 Q44 V31
Posts: 36
20 Aug 2026, 02:31
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n!*(n+1)! = n!*(n+1)*(n!) = n!^2*(n+1). Option E
Bunuel
