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G10

Page history last edited by PBworks 17 years, 1 month ago

Outcome: G10   connect Pascal's Triangle with combinatorial coefficients

 

Originators: Alec & Drew

 

 

Explanation: Pascal's Traingle is an arrangement of binomial coefficients in the shape of a triangle. It is named after Blaise Pascal who created it to demonstrate a method of finding binomial factors.  The rows in Pascal's Triangle are numbered starting with row zero which has a single 1, followed by row 1 which has two 1s. As you go down from row to row, you always carry down a 1 on each end of the row, and you add each number side by side together and place the sum of those numbers underneath and in between those numbers in the row immediately below. I

 

 

 

 

 

[This image is linked from http://mathforum.org/workshops/usi/pascal/images/pascal.hex2.gif ]

 

This trianlge can be connected to combinatorial coefficients because each number in the triangle can be determined by using a combination.  In order to find each value in the triangle you pick the row number that the value you're trying to find is in, and you do a combination by using that number and choosing from that number evey value from 0 to that number. 

 

 

 

Ex: Find all the values in row four of Pascal's Triangle using combinations......    

 

 

  row 4 = 4 C 0, 4 C 1, 4 C 2, 4 C 3, 4 C 4 

 

 

           =  4!/0!*4!, 4!/1!*3!, 4!/2!*2!, 4!/3!*1!, 4!/4!*0!

 

 

           = 1, 4, 6, 4, 1

 

 

 

 

  Ex: Find all the values in row seven of Pascal's Triangle using combinations......

 

 

  row 7 = 7 C 0, 7 C 1, 7 C 2, 7 C 3, 7 C 3, 7 C 4, 7 C 5, 7 C 6, 7 C 7 

 

 

           = 7!/0!*7!, 7!/1!*6!, 7!/2!*5!, 7!/3!*4!, 7!/4!*3!, 7!/5!*2!, 7!/6!*1!, 7!/7!*0!

 

 

           = 1, 7, 21, 35, 35, 21, 7, 1

 

 

 

 

 Now you give it a try!:     

 

Find all the values in the tenth row of Pascal's Triangle using combinations........

 

Solution:

 

  row 12 = 9 C 0, 9 C 1, 9 C 2, 9 C 3, 9 C 4, 9 C 5, 9 C 6, 9 C 7, 9 C 8, 9 C 9

 

             = 9!/0!*9!, 9!/1!*8!, 9!/2!*7!, 9!/3!*6!, 9!/4!*5!, 9!/5!*4!, 9!/6!*3!, 9!/7!*2!, 9!/8!*1!, 9!/9!*0!

 

             = 1, 9, 36, 84, 126, 126, 84, 36, 9, 1

 

 

 

 

 

 

 

Comments (1)

Anonymous said

at 7:46 pm on Jun 14, 2007

Nice work guys. However you need to make your own graphic or give credit to the source. You just poached the graphic from mathforum.org for your page.

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