If ${^nC_9}={^nC_8},$ find ${^nC_{17}}$
Answer & explanation
Correct answer: option 3
\(\frac{ n! }{9! ( n - 9 )!}\) = \(\frac{ n! }{8! ( n - 8 )!}\)
\(\frac{ 1 }{9 ×8 ! ( n - 9 )!}\) = \(\frac{ 1}{8! ( n - 8 )× ( n - 8 )!}\)
n = 17
Now,
${^nC_{17}}$
= ${^17C_{17}}$
= \(\frac{ 17! }{17! ( 17 - 17 )!}\)
= 1
The correct answer is option (3) : 1