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Muchammad Zidan Rachman
Student XII IPA
tentukan nilai n yang memenuhi bentuk permutasi berikut nP4 5040 그
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Putra Gema Nusa
Rockstar Teacher GURU SMA
Pa 5040 n = 5040 TM 14)! -> n + 7 = 0 n=-7 n 7 n - 10 = 0 10 n²-3n+72=0 nER n = . a E oo n =10 = 5040 O JU s s (n- n.(n-1). (n-2). (n-3). fast! (hat! 5040 =(m2-n)(n2-5+6) n²) na-6n²+ 11 n²-6h na-603 + 11n²-6h-5040 o=na +7n3 - 13n² - gin²+ 1020² + 714n-7200-5040 = n3 (n+7)-1302 (n+7)+102n (n+7)-720(n+7) (n+7) (n?-137* + 102n –720) D (n+7)( 13 – 1012 – 3m? + 301 + 721 – 720) Pahamify o = (n +7% (n=10)(n+-3n +72) 0 = (n+7) (12(n-10) – 31 (n-10) + 72 (n–10)) 0 )- 2 0 Đ st j! -
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