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Prove $$\sum_{k=0}^n {n \choose k}=2^n$$ by proving that both sides of the formula represent the number of subsets of a set of $n$ elements. For the left side use the addition rule for counting after partitioning the collection of all subsets according to size. And for the right side use the product rule for counting after identifying a subset $A \subset \{1, 2, ..., n\}$ with the sequence of zeros and ones which is the indicator of $A$.

JMoravitz
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Mathify
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  • what have you done? – Phicar Sep 09 '19 at 19:00
  • What's the question? – Angina Seng Sep 09 '19 at 19:01
  • I do not know what to do for it. It also says that I need to prove it using $$ n \choose k = {n-1} \choose {k-1} + {n-1} \choose k$$. @Phicar – Mathify Sep 09 '19 at 19:07
  • It is asking how to prove the formula using $$ n \choose k = {n-1} \choose {k-1} + {n-1} \choose k$$ too. I do not understand exactly what I am supposed to do @LordSharktheUnknown – Mathify Sep 09 '19 at 19:11
  • As an aside... this is a question about counting, an introductory question about combinatorics. Nowhere in this problem does the concept of probability ever come up. Now... it is possible that this was mentioned in a probability course, but so is addition and multiplication... If the word "probability" doesn't occur in a problem anywhere, you can almost always be sure that it shouldn't be tagged as such. – JMoravitz Sep 09 '19 at 19:31
  • Proving the identity using pascal's identity in your comment can be done by induction like it is done here. – JMoravitz Sep 09 '19 at 19:35
  • As for counting in two ways, this very example is included as one of the examples of this proof technique on the wikipedia page for counting in two ways. The only thing missing from what I closed this as a duplicate of to complete the problem is understanding what counting in two ways means, and recognizing that you can count by breaking into cases based on the size of the subset and adding over all cases. – JMoravitz Sep 09 '19 at 19:39

1 Answers1

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Hint

As a well-known fact, the number of all subsets of a set $A$ with the cardinality $n$ is $2^n$ since each member of $A$ may or may not be in that subset.

As another insight, any subset of $A$ with $k$ members, must be made up of exactly $k$ distinct members of $A$. How then, can you finish the proof?

Mostafa Ayaz
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