Soal Hitungan Dasar Permutasi dan Kombinasi

Soal 1. Diketahui $10 _nP_2= _{n+1}P_4$ Nilai dari $n^2-2n+18=...$

Jawab:
$10 _nP_2= _{n+1}P_4 \\ 10 \frac{n!}{(n-2)!}= \frac {(n+1)!}{(n+1-4)!} \\ 10 \frac{n!}{(n-2)!}= \frac {(n+1)!}{(n-3)!} \\ 10 \frac{n.(n-1)(n-2)!}{(n-2)!}= \frac {(n+1)(n)(n-1)(n-2)(n-3))!}{(n-3)!} \\ 10 =(n+1)(n-2) \\ 5 \times 2 =(n+1)(n-2) \\ n= 4 \\ n^2-2n+18 = 4^2-2.4+18=26$

Soal 2. $_nC_3=2n$ maka nilai dari $_{2n}C_7=...$

Jawab:
$_nC_3=2n  \\ \frac {n!}{(n-3)!3!}=2n \\ \frac {n(n-1)(n-2)(n-3)!}{(n-3)!}=3!.2n \\ (n-1)(n-2)=12 \\ n= 5 \\ _{2n}C_7 = _{10}C_7 $