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For any positive integer n consider the expression $\dbinom{n}{0} + \dbinom{n}{1} 2 + \dbinom{n}{2} 2^2 + ... + \dbinom{n}{n} {2^n}$

Q) use the binomial theorem with appropriate values for x and y to simplify the expression. Now define a set of elements such that when you count the elements in two different ways you get each side of the identity.

Well i am assuming this means $x=y=1$ and that the sum equals $2^{n!} $ a counting argument for $2^{n!} $ (assuming thats what im doing) for $2^{n!} $ seems somewhat simple.

Consider how many different ways i can line up n people then ask them to select a red or blue jelly bean to eat. the other side just looks like gibberish to me.

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    Remember that the binomial theorem states that $(x+y)^n = \binom{n}{0}x^ny^0+\binom{n}{1}x^{n-1}y^1+\dots+\binom{n}{k}x^{n-k}y^k+\dots+\binom{n}{n}x^0y^n$. Note also that $\binom{n}{2}2^2 = \binom{n}{2}\cdot 1\cdot 2^2 = \binom{n}{2}1^{n-2}2^2$ and similarly the others. Do you really want to use $x=y=1$? – 2017-02-05
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    http://math.stackexchange.com/questions/200067/combinatorial-proof-of-an-identity – 2017-02-05
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    Also, as an aside, $2^{n!}$ doesn't count the number of ways you can line up $n$ people and ask them to eat a red or blue jelly bean, the correct count for that would be $2^n\cdot n!$. On the other hand, $2^{n!}$ would be the number of ways to take a subset of the ways to arrange $n$ people in a line. (*E.g. you've written all $n!$ of the different arrangements on pieces of paper, $2^{n!}$ would be the number of collections of paper you could take (order not mattering)*) – 2017-02-05

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Let A and B are two subsets of a set S of n elements. Find the number of tuples $(A,B)$ such that $ A \subset B \subset S$. The upper expression is this one. As choosing a B is choosing $\binom{n}{r}$ elements and then number of A's for that B is $2^r$ and summing over r gives the expression. The other way of counting this situation is that each element x has 3 choices:

a) $ x \in A,B,S$

b) $ x\in B,S ,\not \in A$

c) $ x \not \in A,B \in S$.

Thus the each element has 3 choices .So then total number of tuples is $3^n$. Also the binomal theorem gives the answer $ (1 +2)^n = 3^n$.