Number of unique paths in Pascal's Triangle

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SUMMARY

The discussion focuses on proving that the number of unique paths from entry {0,0} to entry {n,i} in Pascal's Triangle is equal to pascal(n,i), which can be represented as nCi. The base case for n=1 demonstrates that there is one unique path to each of the two elements in the first row. The conversation emphasizes the importance of understanding how to reach the ith element in the nth row while adhering to movement restrictions within the triangle.

PREREQUISITES
  • Understanding of Pascal's Triangle and its properties
  • Familiarity with binomial coefficients, specifically nCi
  • Basic knowledge of mathematical induction
  • Concept of unique paths in combinatorial mathematics
NEXT STEPS
  • Study the properties of binomial coefficients in depth
  • Learn about mathematical induction techniques and their applications
  • Explore combinatorial path counting methods in Pascal's Triangle
  • Investigate movement restrictions in grid-based path problems
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Students studying combinatorics, educators teaching mathematical induction, and anyone interested in the properties of Pascal's Triangle and path counting in combinatorial mathematics.

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Homework Statement


Let pascal(n,i) be the value of the ith element of the nth row of Pascal's triangle. Using induction show that the number of unique paths from entry {0,0} to entry {n,i} in Pascal's triangle is equal to pascal(n,i).

The Attempt at a Solution


The base case n=1 seems easy enough to prove. It is obvious that there is only one path to each of the two elements in the nth row.

It was brought to my attention that each value at {n,i} in Pascal's triangle can be represented by nCi.

That being said I am not sure how to precede. I am just looking for some insight on how to continue.

Any suggestions would be greatly appreciated.
 
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Hint: How (from which elements) can you reach the ith element in the nth row, if movement is restricted in the appropriate way?
 

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