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Given that $^nC_3 -$$^nC_2 = 14$, find $n$. There's a method using the Pascal's Triangle, where the $k^{th}$ entry in the $n^{th}$ row is equal to $^nC_k$. Is there another way to solve this?

user3294195
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1 Answers1

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HINT: if i understand right is this $$\frac{n(n-1)(n-2)}{6}-\frac{n(n-1)}{2}=14$$

Dr. Sonnhard Graubner
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  • Yes, but this would lead to solving a third degree polynomial equation. The Pascal's triangle method is still easier. Looking for a method that is faster than the Pascal's triangle one. – user3294195 Oct 17 '15 at 15:29