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If ${^nC_9}={^nC_8},$ find ${^nC_{17}}$ |
${^{n+1}C_r}$ 10 1 6 |
1 |
\(\frac{ n! }{9! ( n - 9 )!}\) = \(\frac{ n! }{8! ( n - 8 )!}\) \(\frac{ 1 }{9 ×8 ! ( n - 9 )!}\) = \(\frac{ 1}{8! ( n - 8 )× ( n - 8 )!}\) n = 17 Now, ${^nC_{17}}$ = ${^17C_{17}}$ = \(\frac{ 17! }{17! ( 17 - 17 )!}\) = 1 The correct answer is option (3) : 1 |
