NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Pascal's triangle
Lesson

So Pascal's triangle (as we have just seen) has connections to counting techniques and now it is time to formalise these connections. 

Our last activity was this one... 

Activity 8

Write down the following (on one line like I have)

4C0     4C1     4C2     4C3     4C4

Then underneath that write the answers.... 

What connection do you see to Pascal's triangle?

What was the connection? 

You should have discovered that the above answers to the combinations was an entire row from Pascal's triangle. 

The $n$n value is the row number (starting at row $0$0) and the $r$r value is the element in the row, (also starting at $0$0).  

So the value for 9C4, will be the row beginning  1--9..... and be the $5$5th number in the row - (remember we start the element from $0$0).

 

Example

Find the missing elements in the this row from Pascal's Triangle.

$1,9,$1,9, $\editable{A},84,\editable{B},\editable{C},84,36,9,1$A,84,B,C,84,36,9,1

Firstly we know that the lines of the triangle are symmetrical.  This helps us identify  that box $\editable{A}$Ashould be the value of $36$36.  As reading from left to right is the same as reading from right to left. 

This symmetry doesn't help us with the values for $\editable{B}$B or $\editable{C}$C, but we can use our knowledge of combinations to solve this.  

$\editable{B}=\editable{C}$B=C because of of the symmetry.

$\editable{B}$B also equals the value of 9C4 and $\editable{C}$C =9C5, but we also know that 9C4=9C5 (confirming what we already knew from symmetry that the values will be the same).

$\editable{B}$B $=$= 9C4 $=126$=126 

Thus both $\editable{B}$B and $\editable{C}=126$C=126.

 

Pascal's Triangle and the Binomial Theorem

 

$(a+b)^0$(a+b)0 $=$= $1$1
$(a+b)^1$(a+b)1 $=$= $a+b$a+b
$(a+b)^2$(a+b)2 $=$= $a^2+2ab+b^2$a2+2ab+b2
$(a+b)^3$(a+b)3 $=$= $a^3+3a^2b+3ab^2+b^3$a3+3a2b+3ab2+b3
$(a+b)^4$(a+b)4 $=$= $a^4+4a^3b+6a^2b^2+4ab^3+b^4$a4+4a3b+6a2b2+4ab3+b4
$(a+b)^5$(a+b)5 $=$= $a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$a5+5a4b+10a3b2+10a2b3+5ab4+b5

 

Consider the expansions above of $(a+b)^n$(a+b)n.  Particularly note the following patterns.

  • For each expansion to the power $n$n, there are $n+1$n+1 elements.
  • For each term, the sum of the exponents is $n$n.  
  • Powers of $a$a decrease from left to right, from $n$n down to $0$0
  • Powers of $b$b increase from left to right, from $0$0 up to $n$n
  • The coefficients start at $1$1, end at $1$1, AND are the terms of the relevant row from Pascals triangle!

 

Example

What are the coefficients for the expansion of $(a+1)^7$(a+1)7, and then write out the full expansion. 

So we can see that we will have $n+1=8$n+1=8 terms.

We can refer to the relevant row in Pascal's triangle, specifically this row 

This shows us that the coefficients will be 

$1,7,21,35,35,21,7,1$1,7,21,35,35,21,7,1.

Thus the full expansion of $(a+1)^7$(a+1)7 will be

  $\left(a+1\right)^7$(a+1)7
$=$= $a^7+7a^61^1+21a^51^2+35a^41^3+35a^31^4+21a^21^5+7a^11^6+1^7$a7+7a611+21a512+35a413+35a314+21a215+7a116+17
$=$= $a^7+7a^6+21a^5+35a^4+35a^3+21a^2+7a+1$a7+7a6+21a5+35a4+35a3+21a2+7a+1

Worked Examples

QUESTION 1

You are given some of the entries in a particular row of Pascal’s triangle. Fill in the missing entries.

  1. $1$1 , $8$8 , $\editable{}$ , $56$56 , $70$70 , $\editable{}$ , $28$28 , $\editable{}$ , $1$1

QUESTION 2

How many terms are there in the expansion of $\left(m+y\right)^8$(m+y)8?

QUESTION 3

Using the relevant row of Pascal’s triangle, determine the coefficient of each term in the expansion of $\left(5+b\right)^5$(5+b)5.

  1. $\left(5+b\right)^5$(5+b)5$=$=$\editable{}$$\times$×$5^5b^0$55b0$+$+$\editable{}$$\times$×$5^4b^1$54b1$+$+$\editable{}$$\times$×$5^3b^2$53b2$+$+$\editable{}$$\times$×$5^2b^3$52b3$+$+$\editable{}$$\times$×$5^1b^4$51b4$+$+$\editable{}$$\times$×$5^0b^5$50b5.

Outcomes

M8-3

Use permutations and combinations

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