I need help in my mathematical reasoning class!

[abaker]

the question that I CANNOT figure out is this:
explain why 2^n = (n choose 0) + (n choose 1) + (n choose 2) + ... + (n choose n-1) + (n choose n)

so far my work is:
(n choose 0) = n!/0!n= 1
(n choose 1) = n
(n choose 2) = (-n)2/2!=1/2(n-1)n
(n choose n-1) = n!/1!(-1+n)! = n
(n choose n) = n!/0!n = 1
I don't know if I am approaching this the right way or how to approach this kind of question

thanks for any help!

 

[tkhunny]

Where are you going with that?

What's (x+1)^n expanded?

After doing that, what happens when we pick x = 1?

 

[soroban]

Hello, abaker!

Explain why: .\(\displaystyle 2^n \:=\: {n\choose0} + {n\choose1} + {n\choose2} + \hdots + {n\choose n-1} + {n\choose n}\)

Since it said "explain", I assume they will accept an explanation.


There are \(\displaystyle n\) objects.

In how many ways can you make a selection
. . (some, all, or none of the objects)?



One explanation

For each of the \(\displaystyle n\) objects, there are two choices: .take it or leave it.

Therefore, there are: .\(\displaystyle 2^n\) possible outcomes.



Another explanation

List and count the possible outcomes.

. . \(\displaystyle \begin{array}{ccc}\text{Take 0 objects:} & {n\choose0}\text{ ways} \\ \\ \text{Take 1 object:} & {n\choose1}\text{ ways} \\ \\ \text{Take 2 objects:} & {n\choose2}\text{ ways} \\ \vdots & \vdots \\ \text{Take }n\text{ objects:} & {n\choose n}\text{ ways} \end{array}\)


Therefore, there are: .\(\displaystyle {n\choose0} + {n\choose1} + {n\choose2} + \hdots + {n\choose n}\) possible outcomes.