Question
If $\frac{1}{{ }^{4} C_{n}}=\frac{1}{{ }^{3} C_{n}}+\frac{1}{{ }^{6} C_{n}}$, then value of $n$ is(a) 3(b) 4(c)(d). 2
Step 1
We can rewrite this equation in terms of factorials as $\frac{1}{\frac{n!}{4!(n-4)!}} = \frac{1}{\frac{n!}{3!(n-3)!}} + \frac{1}{\frac{n!}{6!(n-6)!}}$. Show more…
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