Question

if, nC12 = nC18, then find nC3?

Answer 1

please find the solution in the attached image

Answer 2

$Given \: ^{n}C_{12} = ^{n}C_{18}$

$We \: know \: that,$

$\pink { If \: ^{n}C_{p} = ^{n}C_{q} \:then }\\\pink { either } \blue {\:\:p=q \:or \: p+q = n }$

Case 1:

$p = q \\12 = 18 \\Which \:is \:not \: possible$

Case 2:

$p + q = n$

$\implies 12 + 18 = n$

$\implies n = 30$

$Now, we \:have \:to \:find$

$^{n}C_{3} \\= ^{30}C_{3}\\=\frac{30!}{(30-3)! 3!}\\= \frac{30!}{27! 3! } \\= \frac{30\times 29 \times 28 \times 27! }{27!\times 6 } \\= 5 \times 29 \times 28 \\= 4060$

Therefore.,

$\red { Value \: of \: ^{n}C_{3} } \green {= 4060}$

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