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Problem: Use the Binomial Formula to show that if $n$ and $r$ are integers with $0 \leq r \leq n$, then $\binom{n}{r}=\binom{n}{n-r}$.

My attempt: I am using the general binomial expansion formula to establish the following.

$(n+r)^{n}=n^{n}+nn^{n-1}r + ...$

But am not sure where to go from here. Should I do a proof by induction with base case $n=0$? I have a feeling induction is not necessary and this problem is easier than I think...

Edit: This question is different than the proposed duplicate because the "duplicate" does not use the binomial theorem method, which I must use. It uses a different method.

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    Possible duplicate of [Proving that ${n}\choose{k}$ $=$ ${n}\choose{n-k}$](http://math.stackexchange.com/questions/404417/proving-that-n-choosek-n-choosen-k) – 2017-01-23
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    This is not a duplicate, at least of that question. OP specified that Binomial Formula must be used. And I don't understand the downvotes. OP has written tries and thoughts. Just because a question is easy it needn't be downvoted. – 2017-01-23

2 Answers 2

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Hint:

The polynomials $(x+y)^n$ and $(y+x)^n$ are the same. Expand and equal their coefficients.

Full answer:

The polynomials $(x+y)^n$ and $(y+x)^n$ are equal. If we apply the Binomial Formula to both of them, we obtain $$\sum_{k=0}^n\binom nk x^ky^{n-k}=\sum_{j=0}^n\binom nj y^jx^{n-j}$$ Fix a degree $r$ for the variable $y$. The term that has this degree in the LHS is $$\binom n {n-r}x^{n-r}y^r$$ and in RHS we find $$\binom nr y^rx^{n-r}$$ This implies $$\binom nr=\binom n{n-r}$$

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    Hi, thanks for the input. I'm not exactly sure how your 2 polynomials there answer the question (but i agree they are equal...) But when I expanded out the coeefficients for my problem, I get: $\sum (\binom {n}{k})n^{n-k}r^{k}$. Is it supposed to be equal if I do it also for the expansion of (n+n-r)^{n}? – 2017-01-23
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    Yes now it makes 100% sense now. I was complicating this problem more than i needed to, I think. Thank you. – 2017-01-25
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In the binomial expansion of $(x+a)^n $, multiplying together $n $ factors, $(x+a_1), (x+a_2), \cdots (x+a_n) $, the indices of the factors of each term added together equals $n $. The coefficient of the term $x^r $ consists of terms $a_1, a_2, \cdots $ such that the sum of indices in each term is $n-r$, and since there is no reason why one combination of these letters should occur more than another, it will consist of all terms satisfying the condition that sum of indices equals $n-r $.

If all $a_i$'s are equal, then each of the mentioned terms becomes $a^{n-r} $, and the number of them will be $\binom{n}{n-r} $, since it is obvious that this is the number of ways we can form $n $ letter combinations of the type described.

Hence in $(x+a)^n $, the term involving $x^r$ is $\binom {n}{n-r}x^ra^{n-r} $ or $\binom {n}{r} x^ra^{n-r} $. Hope it helps.

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    Can you please explain how we know that $\binom{n}{n-r}x^{r}a^{n-r}$ is equal to $\binom{n}{r}x^{r}a^{n-r}$? – 2017-01-23
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    I got the answer now - I gave another user the "accepted" answer for being more complete, but your input helped lead me to the right idea. Thanks very much. – 2017-01-25