Permutations and Combinations

3.4 More Patterns

Lesson materials located below the video overview.

P24001

Exercise 33: Notice:

\(1 = 1\)

\(1 + 1 = 2\)

\(1 + 2 + 1 = 4\)

\(1 + 3 + 3 + 1 = 8\)

\(1 + 4 + 6 + 4 + 1 = 16\)

a) Without doing the computation, explain why the sum of entries in any row shown above will turn out to be double the sum of the entries in the row just above it.

HINT: Each entry in any one row is the sum of two entries in the row above it.

 

b) Explain why the sum of entries in the \(n\)th row is sure to be \(2^{n}\). (Remember to call the top row of Pascal’s triangle the zero-th row.)

 

Exercise 34: Explain why each alternating sum in Pascal’s triangle, beyond the zero-th row, is zero:

\(1 – 1 = 0\)

\(1 – 2 + 1 = 0\)

\(1 – 3 + 3 – 1 = 0\)

\(1 – 4 + 6 – 4 + 1 = 0\)

\(1 – 5 + 10 – 10 + 5 – 1 = 0\)

 

The following property is strange. Look at the powers of \(11\):

\(11^{0} = 1\)

\(11^{1} = 11 \)

\(11^{2} = 121 \)

\(11^{3} = 1331 \)

\(11^{4} = 14641\)

\(11^{5} = 161051 = 1|5|10|10|5|1\) (See Exploding Dots!)

 

Any guesses as to why these powers appear as rows of Pascal’s triangle?

 

 

 

 

 

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