Instructions for Working with Pascal's Triangle and Fractals

Rachel's coloring of multiples of 5                               Jessica's coloring of multiples of 3

When teaching Probability

By 8th grade students are able to understand the probability of an event.  They've done experiments in lower grades with the probability of coin tosses.  I like to delve into this same topic again in 8th grade by asking students this question,

"If I toss these 3 pennies, what is the probability that I will get 2 heads and 1 tail?"

After some stumbling we decide that we have to know how many different ways the coins could land and how many ways the coins could land as 2H + 1T. 

The probability would then be successes divided by possible ways.  We start listing.

HHH           HHT     When I say "HTH", some students say that is the same as HHT. 

Then we discuss whether it matters if it is the shiny penny or the dull one that has tails showing.  We decide that these are two different events ... HHT and HTH.

Back to counting;

HHH         HHT       HTH       THH         HTT      THT        TTH           TTT

So what is the probability of getting 2H and 1 T.  We decide that there was only one way of getting all heads or all tails and three ways of getting just one head or just one tail.  Therefore there were 8 possible outcomes to the toss.  Three ways of getting 2 heads and one tail out of 8 ways of tossing the coins then the probability of tossing 3 coins and having 2H and 1T be the outcome is 3/8.

I might ask "What if I toss 4 coins?  What would the probability of getting 2H and 2T be?  We list these arrangements.      

HHHH 
HHHT
HHTT
TTTH
TTTT
HHTH
HTTH
TTHT
HTHH
TTHH
THTT
THHH
THTH
HTTT
HTHT
THHT

We are able to agree that there are 16 ways of tossing the coins and 6 of those ways are 2 T + 2 H.  Therefore the probability of getting 2T and 2H is not 50% as was expected by some but 6/16 or 3/8.

Ready for more groaning?  What if I tossed 5 coins?

OK.  Maybe there is another way.

 

Beginning of Pascal's Triangle lesson:

There was an excellent mathematician named Blaise Pascal who lived in France around 1650, who showed this interesting arrangement of numbers.

1
1      1
1      2      1
1      3       3     1
1     4      6      4       1
1     5      10     10     5      1

As a class we discuss what student's can see in the pattern ...

Can you see how one row is calculated from the previous row?

So why am I showing you Pascal's triangle during this lesson on probability.  Students recognize that our 3-coin situation had;

1 way of having 3H;    3 ways of having 2H-1T;    3 ways of having 1H-2T;    and 1 way of getting 3T.

The number of ways for each possibility is just like the 3rd row of Pascal's triangle ... 1    3   3   1

They find that for four coins the same pattern of         1          4          6          4          1           had been useful.  To find the probability of getting 4 H, you would add the elements of the 4th row of Pascal's triangle = 16.  That would show the possible number of combinations.   There is only 1 way to get 4H (1st number in the row).  Therefore the probability of getting 4 heads is 1/16. 

.

Messing around with Pascal's triangle - class work and first night of homework.

Pascal's triangle turns out to be symmetrical.  Ask the class why that is?  This observation is useful for students to use when checking their addition for accuracy.

I give students a grid like the one below.  There are better grids made with hexagons.

In class I ask students to complete the table without calculators and color in the multiples of one number. It's fun to assign the multiples of 2, 3, or 5 to students who seem to like coloring. Multiples of 6, 7, 11, or 13 can be assigned to students who might enjoy the challenge of finding more obscure multiples.  Each student should be assigned one multiple to color.  Students will see triangles emerge from their colorings that have the triangle vertex point facing downward.  Ask why they suppose that this happens?

Example:

               1               
1
1
1
2
1
1
3
3
1
 1 
4
6
4
 1 
1
5
10
10
5
 1 
 1 
6
15
20
15
6
 1 
 1 
 7 
21
35
35
21
7
 1 

This exercise was practice with addition and carrying in your head. The sum of two multiples of 3 must be a multiple of 3, why?   

Answer:  1 three + 1 three = 2 threes
            2 threes + 5 threes = 7 threes       

This is another way of recognizing the validity of the distributive property,  dist-property-equation   

 Pascal's Triangle with multiples of 2 colored                                       Pascal's Triangle with multiples of 7 colored

 

Second day in class

Filling in Pascal's triangle quickly involves large numbers. The numbers are cumbersome to write and motivate students to start using calculators instead of their heads ... mistakes then happen.  For finding the multiples of 7 (for instance) dividing by 7 to see if these numbers are multiples of 7 seems way too hard.  So, here students will have motivation to learn modular math.

This is a modular math table for adding two numbers mod 7.

+ 0 1 2 3 4 5 6
0 0 1 2 3 4 5 6
1 1 2 3 4 5 6 0
2 2 3 4 5 6 0 1
3 3 4 5 6 0 1 2
4 4 5 6 0 1 2 3
5 5 6 0 1 2 3 4
6 6 0 1 2 3 4 5

Enjoy a class discussion of how this works and some practice work.  Don't clocks work just like this?  Clocks are in mod what?

Now when students are given a new number to find multiples of in Pascal's triangle the addition will be much easier.  In our previous example, a multiple of 3 triangle would look like this.

              1              
1
1
1
 2
1
 1 
0
0
 1
 1 
 1 
0
 1
 1
 1 
2
 1
 1
2
 1
 1
0
0
2
0
0
 1
 1
 1
0
2
2
0
1
 1

Give a second night of coloring and multiple homework with smaller grid paper (since now student's generally have to only write 1 digit numbers).

There will be students who really enjoy this task and want to make incredibly large multiple pictures.  Taping grid papers together allows students to continue to any extent that they wish.

 

When teaching the Binomial Expansion

I always find it surprising that students who are adept and confident with the distributive property don't initially see the issues with evaluating any numbers of the form (x+y)squared.  Too often students will just assume that this expression means xsquared+ysquared.  Therefore, it seems imperative to me that students spend some time in an investigation of binomial expansions.

Generally I begin with trying to evaluate (x+1)squared.  To convince students that their gut reaction to simplifying this expression is not xsquared +1, I suggest that we see what would happen if we knew the value of  x.  So, we might assume that x = 6.  Then surely (6+1)squared should equal 49.  But is 6squared+1squared=49?  Whoops.  What happened?  From this example, students generally suggest some form of double distribution and we are on our way.

Now to make continued use of the distributive property as we multiply sums, we belabor this issue of binomial expansion further.

            series of squared binomails

The next question for the class is what might (x+1)cubed turn out to mean.  By now students seem comfident that they have to use the distributive property to distribute each member of the sum.  They generally willingly do this example.

But, with this question, (x+3)to the fourth they begin to revolt.

This is where it is fun to show Pascal's triangle.  Return here to the section above marked Beginning of Pascal's Triangle lesson.  The presentation and application of Pascal's triangle included in the probability section is the same as what you would use for playing with Pascal's triangle as a tool in binomial expansions.

Enjoy!