[ (n+4)! = 3024 \cdot n! \implies (n+4)(n+3)(n+2)(n+1) n! = 3024 \cdot n! ] [ (n+4)(n+3)(n+2)(n+1) = 3024. ]
Try ( n=5 ): ( 9\cdot8\cdot7\cdot6 = 3024 ). Yes.
Answer: ( n = 5 )
11TH•Mathematics•Permutations And Combinations•Question 3
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Find $n$, if $(n+4)! = 3024 \cdot n!$.
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