Know your tricks!

Sum of elements = $2^{n}$

In this case, $n = 20$

$2^{20}$ = $1048576$

METHOD 1: Consider the expansion of $\displaystyle (1+x)^{n} = \dbinom{n}{0} + \dbinom{n}{1} +...+ \dbinom{n}{n-1}x^{n-1} + \dbinom{n}{n}x^{n}$

When x=1, n=20

$\displaystyle \dbinom{20}{0} + \dbinom{20}{1} +...+ \dbinom{20}{19}+ \dbinom{20}{20} = \boxed{2^{20}}$

METHOD 2: Consider twenty boxes and twenty toys which are to be distributed among them, with the constraint that each box can contain at most 1 toy. Then we can choose the boxes in which we place the toys in ${20 \choose 0}+ {20 \choose 1} + {20 \choose 2} + {20 \choose 3} + .... + {20 \choose 19} + {20 \choose 20}$ ways. Now, for any given box, the toy can be either in it or not, giving us 2 possibilities and by the multiplicative principle, for 20 boxes, the possibilities become $2^{20}$ . Also, the number of possibilities arrived at by two different ways must be equal(assuming we reasoned correctly)

Hence, ${20 \choose 0}+ {20 \choose 1} + {20 \choose 2} + {20 \choose 3} + .... + {20 \choose 19} + {20 \choose 20} = \boxed{2^{20}}$