Solution to the Path Problem
Start A Row 1 B C Row 2 D E F Row 3 G H I J
Consider the number of paths to the letters in Row 1:
There is only 1 path from A to B
There is only 1 path from A to C
Next consider the number of paths from A to a letter in Row 2:
There is only 1 path from A to D:
A - B - D
There are 2 paths from A to E:
A - B - E and A - C - E
There is only 1 path from A to F:
A - C - F
The number of paths in these two cases corresponds to the first two lines in Pascal's Triangle. Consider now the paths that start with A and lead to a letter in Row 3:
There is only 1 path from A to G:
A - B - D - G
There are 3 paths from A to H:
A - B - E - H
A - C - E - H
A - B - D - H
There are 3 paths from A to I:
A - B - E - I
A - C - E - I
A - B - F - I
There is only 1 path from A to J:
A - C - F - J
The number of paths to a letter in Row 3 corresponds to the entries in the third row of pascal's triangle.
Some Conclusions About Using Patterns in Pascal's Triangle To Solve Problems
In both problem 1 and problem 2, at each step (a flip for the coins, a move downward for the paths), there are two possible outcomes. With the coin we either get a Head or a Tail. With the paths we can either go Left or Right.
Start A Row 1 B C Row 2 D E F Row 3 G H I J Note that each path to a particular letter has the same number of left and right choices, though in a different order.
Consider the paths from A to H:
Path Description Lefts and Rights A - B - D - H Go left twice then right once L - L - R A - B - E - H Go left then right then left L - R - L A - C - E - H Go right once then left twice R - L - L Each path consists of going Left twice and Right once.
Conclusion
Whenever a problem involves two equally likely outcomes at each step (Heads or Tails; Left or Right), the possible results after n steps will be the nth row of Pascal's Triangle.
The first number (and the last number) in the row, always a 1, will represent getting n identical results (all Heads or all Tails; all Lefts or all Rights). The next number will represent getting (n - 1) of 1 results and 1 of the other result, etc.
Thus, any problem that can be interpreted this way, can be solved by using the entries in Pascal's Triangle.
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email the author: Bruce Jacobs
Last modified: Thursday, February 6, 2003