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Binomial Theorem Formula:

  • Sigma Notation: $\sum_{i=0}^{n} $ $n\choose i $$ x^i $ $y^{(n-i)}$

Note: $(n-i)$ is the exponent of $y$.

Logical Question:

  • Can the place of exponents for $x$ and $y$ be switched ?

Ex. $\sum_{i=0}^{n} $ $n\choose i $$ x^{(n-i)} $ $y^i$

Note: Now $(n-i)$ is the exponent of $x$.

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    you should use braces to obtain $x^{n-i}$ instead of $x^(n-i)$ – 2017-02-18
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    The key for your question is the symmetry of binomial coefficients : $\binom nk=\binom n{n-k}$ – 2017-02-18
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    Ok. Yes, that was the key to this problem. Thank you! I believe you should post the second comment as the answer to this question and I will accept it and upvote it for you. – 2017-02-18
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    `x^{(n-i)}` instead of `x^(n-i)` – 2017-02-18
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    Yes indeed. Fixed it now. 2 other people agreed that it is possible to switch the exponents of i and n - i for x and y. Try n choose k and then n choose n - k. Substitute the exponents and you'll see the exponents of x and y can be changed. – 2017-02-18

2 Answers 2

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Hint:

Note that $$(x+y)^n = (y+x)^n$$

The general term of the first expression is $\binom {n}{r}x^ry^{n-r } $ and in the second case is $\binom{n}{r}y^rx^{n-r} = \binom {n}{n-r}x^ry^{n-r} $. Thus, both expressions are equivalent. Hope it helps.

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    Ok, so we can switch the place of exponents since for addition and multiplication there is the option of replacement. Thus, it indicates exponents can be switched as well ? – 2017-02-18
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    Ok, thank you for your help. – 2017-02-18
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The key for your question is the symmetry of binomial coefficients ... for all integers $n,k$ such that $0\le k\le n$ we have :

$$\binom nk=\binom n{n-k}$$

This can be understood with a combinatorial argument : given a set $E$ such that $\mathrm{card}(E)=n$ and an integer $k$ such that $0\le k\le n$, there exists a bijection from the set $\mathcal{P}_k(E)$ of subsets of $A\subset E$ such that $\mathrm{card}(A)=k$ to the set $\mathcal{P}_{n-k}(E)$ : map $A$ to $E-A$.

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    0 ≤ k ≤ n is another important key point. – 2017-02-18