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Question:

Can binomial coefficient be written in both ways?

$0 \leq k \leq n$

$\therefore \dbinom{n}{k} = \dbinom{n}{n-k}$

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    Somebody told me that formula, but when I am searching I don't see such a thing, and I am starting to doubt t. Could you please provide some explanation and confirmation ? – 2017-02-18
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    why are you in doubt? – 2017-02-18
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    I mean does this n choose k = n choose n - k exist ? If yes, does it mean that the place of exponents for x and y can be switched ? For example, $x^k * y^{n-k}$ vs $x^{n-k} * y^k $? – 2017-02-18
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    What it means is that the binomial coefficients read the same forwards and backwards. Choosing $k$ to leave in is the same as choosing $n - k$ to leave out. – 2017-02-18
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    The short answer to your question is yes. Have you tried writing both sides of the equation in terms of factorials? – 2017-02-18
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    @N. F. Taussig , you mean developing them ? Yes, I got this $\therefore$ $n^k * y^{n-k}$ = $n^{n-k} * y^k$ – 2017-02-19
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    What I meant is that you can show that $$\binom{n}{k} = \binom{n}{n - k}$$ by applying the formula $$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$ to obtain $$\binom{n}{n - k} = \frac{n!}{(n - k)![n - (n - k)]!} = \frac{n!}{(n - k)!k!} = \frac{n!}{k!(n - k)!} = \binom{n}{k}$$ – 2017-02-19
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    @N.F.Taussig , nice one my friend. :) – 2017-02-19

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The binomial coefficient ${n \choose k}$ is the number of ways to choose a group of $k$ elements from a set of size $n$. When you choose a group of $k$ elements, there are $n - k$ elements left unchosen. So, instead of choosing $k$ elements to be included, you can choose $n-k$ elements to not be included, and it's the same thing. So that's why ${n \choose k} = {n \choose n - k}$