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I'm trying to proof the symmetric identity ${n \choose k} = {n \choose n-k}$ using induction.

I have the base case $${0 \choose 0} = 1 = {0 \choose 0-0}$$ for k=0 and $${0 \choose k} = 0 = {0 \choose n-k}$$ for every other k>0. My hypothesis is ${n \choose k} = {n \choose n-k}$ for any natural number, and I need to show that ${n+1 \choose k} = {n+1 \choose n-k+1}$. I used Pascal's Identity to show that $${n+1 \choose k} = {n \choose k} + {n \choose k-1}$$ which is equal to $${n! \over (n-k)!k!} = {n! \over (n-k+1)!(k-1)!} $$. At this point I don't know how to go on. I am generally not sure if I have done the induction steps correctly (I am a biology student), and would be very appreciating of any help.

3 Answers 3

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As k99731 has pointed out, this result follows immediately from the definition.

Instead consider the following recurrence relation $A_{n,k}$ where we define $$ A_{n,k} = A_{n-1,k} + A_{n-1,k-1}$$ for all integers $n$ and integers $k$ such that $1\leq k\leq n$ and set the boundary conditions $A_{n,n}=A_{n,0}=1$ for every integer $n$.

Can you use induction to prove that $A_{n,k}=A_{n,n-k}$ for all $n,k$?

Hint: Let $S(n)$ be the statement that $A_{n,k}=A_{n,n-k}$ for all $1\leq k\leq n$. Can you prove that $S(n+1)$ will be true if $S(n)$ is true?

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${n \choose k}= \frac{n!}{(n-k)! k!}=\frac{n!}{k! (n-k)!}={n\choose {n-k}}$

Did I misunderstand the purpose of this question?

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    I am trying to do an inductive proof for which I am asking for help. – 2017-01-03
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    @Quebec27 Well you can write the above sentence after you assume the induction hypothesis... I would say it is really a bad idea to write an inductive proof, since the induction hypothesis will not help you. Especially when you write the formula for ${n\choose k}$ out, the conclusion becomes obvious. In case you want some exercise, try to prove $1^2+2^2+\dots+n^2=\frac{n(n+1)}{2}$, by induction. – 2017-01-03
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    Yeah, I wanted to see for myself how induction can be used in not so "convenient" cases. Thanks anyway. – 2017-01-03
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It is more complicated than necessary, but your induction argument can be completed. Starting from $$\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}$$ apply the induction hypothesis to both terms on the right side (remember the induction hypothesis applies when the top is $n$ with any value on the bottom), which gives $$\binom{n+1}{k} = \binom{n}{n-k} + \binom{n}{n-k+1}$$ Now apply the Pascal identity once more on the right side, to get what you want: $$\binom{n+1}{k} = \binom{n+1}{n-k+1}.$$

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    Okay thank you for your helpful response! One more thing, is the base case with ${0 \choose k} = {0 \choose 0-k}$ for any k>0 correct? – 2017-01-03
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    @Quebec27 Yes. In fact, for nonnegative $n$, if we define $\binom{n}{k} = 0$ for all integers $k<0$ or $k>n$, then Pascal's identity will still hold. Just imagine extending Pascal's triangle with 0's indefinitely to the left and right on each row, and check that Pascal's identity still works. – 2017-01-04