Pascal’s Triangle is a fascinating concept.

http://ptri1.tripod.com/
After extending the triangle and viewing the sums of the numbers in the rows, I found it to be a geometric sequence. ( 2, 4, 8, 16, 32)
The activities dealing with the probability of teenage accidents would be one to share with those who are preparing to get their driver’s licenses!
3. Number of teenagers having an accident 0 1 2 3 4 5
Number of ways 1 5 10 10 5 1
4. The probability that none of the five teenagers having an accident is 0.65 = 0.07776 or 8%.
5. The probability that exactly one teenager can have an accident is 5 x (0.4 x 0.64) = 0.2592 or 26%.
6-7. There are 10 ways exactly two teenagers can have accidents so the probability would be 10 x (0.42 x 0.63) = 0.3456 or 35%.
8-9. There are 10 ways exactly three teenagers can have accidents so the probability would be 23%.10 x (0.43 x 0.62 = 0.02323)
10-11. There are exactly 5 ways exactly four teenagers can have accidents so the probability would be 7.7%. 5 x ( 0.44 x 0.6 = 0.0768)
12.The probability that all five teenagers will have accidents is 1%. (0.45 = 0.01024).
Using the triangle for solving problems that would be relevant for elementary children can be a challenge. I wanted to find some elementary examples that I could use as models for developing some good questions. I found a simple, clear example:
http://www.krysstal.com/binomial.html
This example involve choosing 2 books out of a possible 5. It showed how to line up number of selections and number of ways like we explored in our lesson.
Another good way to find an answer to a “line item” was found at http://ualr.edu/lasmoller/pascalstriangle.html:
“One can in fact find the number of ways of choosing k items from a set of n items simply by looking at the kth entry on the nth row of the triangle. So, to see how many different trios you could form using the 45 members of your jazz band, you would look at the 3rd entry on the 45th row. (The “1″ at the top of the triangle is considered the “0″th row, and the first entry on each row is labeled the “0″th entry on the row.
So let’s see if this can work with choosing letters from one’s name. If my name is Kris and I wanted to choose two letters, how many ways could be chosen?
Using the table: 0 1 2 3 4
1 4 6 4 1
I should get 6 different ways: KR KI KS RI RS IS
(other combinations would be the same letters, just reversed)
If I use the “kth entry on the nth row of the triangle” I would also get 6
[the 2nd entry of the fourth row (1 4 6 4 1)]
I think students would have fun figuring out the letter combinations using their own names!
Another good question they would enjoy solving would be choosing a number of different treats from a list of eight to create Halloween Goodie Bags. For example:
You have 8 different types of candy and you want to put 3 different treats in each goodie bag, how many different goodie bags would be made?
Using the table 0 1 2 3 4 5 6 7 8
1 8 28 56 70 56 28 8 1
or the triangle row: 1 8 28 56 70 56 28 8 1
Using this model, they could create their own problems.
At my grade level (grades 3-5) I wouldn’t go into the probability extension like we processed in the lesson. If they can do examples like the ones above, they will be prepared for figuring out the binomial probability later in middle and high school.