Monday, December 1, 2008

Just Another SAT-Type Combinatorial Problem

How many 3-digit positive even integers have at least one digit equal to 2?

(1) Talk about intense information overloading! Students need considerable exposure to these short but concentrated exercises. There are at least 6 pieces of information packed into these 14 words! What % of students do you think would miss or misinterpret one or more of these 'clues'?
(2) Do they really put questions like this on the SATs? Ask any student who recently took the PSAT!
(3) There are many strategies possible here. Many students (if they fully comprehend the question) will start listing 102,112,120,... but how successful will they be using this approach? There is a powerful approach for the "at least one" types of counting problems. I strongly advocate this starting in middle school.
(4) I have published many other similar problems (look under combinatorial math in the index). Do these get easier with practice over time? I think so but there always needs to be clarity of thinking and a careful organized approach. The quick clever student often falters when detail is required. This may help that student to mature!
(5) Are you thinking I'm making too big a deal over SAT-types of questions? What if your students won't even take these tests, choosing ACTs instead? Hopefully, you will come to believe that my purpose is to use these kinds of problems simply as a vehicle for taking students to a higher level of thought. What would be the harm of using these for the occasional class opener (aka Warm-Up, Problem of the Day, Do Now, etc.). In fact I would encourage this at least once a week!

Another compelling reason to discuss more than one method of solution for combinatorial problems: One is rarely 100% certain of the accuracy of one's answer without doing the problem by an alternate method and getting the same result!


jd2718 said...

These look so different from my perspective... teaching a high school course in this stuff.

I would expect a few kids to list 102, 112, 122, 132, 142, 152, 162, 172, 182, 192, 120, 124, 126, 128

14*8 = 112

+ 200, 202, ... 298

+ 50 = 162

Those who grasp the stuff better will have work that looks like:

900/2 - 8*9*4
(half the 3 digit numbers are even, - 288 of those include no 2s)

I really think that the topic or group of topics improves problem solving skills elsewhere...


Dave Marain said...

Thanks, Jonathan.
I concur with both your answer and your assessment of how many students would list, but we don't know for sure unless we try it out. However, I have observed many students, both in middle school and high school, listing in simple numerical order.

Our goal should be to promote development of more efficient systematic approaches as students progress through the grades but, sadly, this is not what I see happening.

Middle school, IMO, is a time for exploration and discovery as well as attainment of important foundation skills and knowledge.

As always, I'm appreciative of your insight and your commitment to the importance of these kinds of problems. The ease with which you solved this problem belies the fact that, for the vast majority of learners, numerous problems of this type (and other types) must be practiced before one achieves competency and confidence!

It always brings me back to the aphorism: An expert is one who has experience!

mathmom said...

I'm going to try assigning this problem for homework with my middle school group later this week. I think you're spot on with the addendum! I almost always try to do combinatorics problems more than one way to convince myself that I got it right (or to find out that I didn't!)

I don't know if I have anyone who will do anything more than listing the possibilities, but my instructions explicitly state "Try to check your answer by solving the problem a second way!" We'll see. I'm also going to have them work on the remainders challenge in class Wednesday. I'll let you know how it goes.