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I'm trying to solve the following equation,

$$3\cdot\binom{n}{k} = \binom{n}{k+1}$$

I'm having trouble on how to work this 3 in.

This relates to Pascal's triangle where the next spot over (For the right half of the triangle) is 3 times less than the previous one.

Try not to give me the answer straight out but just enough to go off of, I should be able to figure the rest out!

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    Do you know a formula (involving factorials) for the entries in Pascal's triangle? – 2017-02-16

2 Answers 2

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$$ 3\cdot {{n}\choose{k}}=3\cdot\frac{n!}{k!(n-k)!}=3\cdot\frac{n!}{(k+1)!(n-k-1)!}\cdot\frac{k+1}{n-k}={n\choose k+1}\cdot\frac{3(k+1)}{n-k} $$ Hence $$ \frac{3(k+1)}{n-k}=1 $$ You should be able to find $k$ as a function of $n$ (ore vice versa) from here

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HINT:

Do you know that $$\binom{n}{k}=\frac{n!}{k!(n-k)!}$$ How about substituting this in that equation of yours?

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    I have, but it still has me perplexed – 2017-02-16