Wednesday, May 13, 2009

Problem Solving - Combinations

Hey guys Ingrid here today to scribe for y'all!

Anyway, today we tried to finish off the unit so that we could write the pre-test tomorrow and then the unit test on Friday, however, we were not able to do so. Mr. K considered moving the test to Tuesday... BUT, the last couple of slides pertaining to the game of poker is for homework. So yes, you guys are still writing the pre-test tomorrow, and the unit test on Friday :).

(Oh yeah, get those deliciosos links in as well ;D)


Wow, everyone's pretty much bumped my scribe post down with their BOB posts, but hey that's a good thing - you guys are listening! Haha.
For those of you who need help for today's homework from the slides... Aldrin included them on his most recent post! Thanks bro!

Okay, now to get to scribing~ Please forgive my lack of explanation skills!

Firstly, for this specific post, let's keep the choose formula in mind. That is;

We immediately started off with some exercises...
Alright, here we were asked to find the 4th term in the given expansion. In order to do so;
  • We let a =
  • and b = or
  • We then use the choose formula and the binomial theorem.
  • Remember that n is the number of terms in the expansion
  • and that r is the term we are looking for... BUT it is always one less than what we are looking for, keep in mind the first term is to the power of 0 and the second term is to the power of 1, etc etc... Does that make sense? >_<
  • Oh, the exponents on the variables always equal to the degree on the expansion.
  • So since r is the exponent on the variable b, it is then easy to see that the exponent on the variable a is 4.
  • 3 + 4 = 7!
  • Mmk from there we just substitute everything in...
  • and solve!

NEXT!This one was an interesting problem. All we really need to know is that
  • There is no middle term when the expansion has an even number of terms!
  • This is because there is no zeroth term, even in University!
  • However, we did talk about the zeroth row.
  • BUT, we don't deal with that in HS either.

  • For this question, there were three different solutions, however only one was correct!
  • Thanks Alex, PJ, and Jessica for your clever ideas!
  • It turns out, that Alex's (the first slide) was the correct method.
  • This is because his method took into account all three possibilities without making it into a permutation.
  • The other two methods on the other hand, were permutations, not combinations.
  • Firstly, Alex took into account that only one boy is in the committee, and the other two spots are girls. So that makes 12 choices for the boy, and 10 girls for the last two spots.
  • Here, he used the choose formula then multiplied that answer by the number of choices for boys (12)!
  • Then he did a similar thing for the next set, which is two boys and one girl..
  • For the last set, he took into account that all three spots are taken by boys.
  • For the final result, he added all three sets!
  • NEXT!
  • Here we have another interesting problem...
  • There are nine chairs in a row but four of these nine students must sit in a consecutive order!
  • First, we figure out that there are 6 possible ways to seat a group of four people with nine chairs.
  • So we placed those four people in a bag, but now we let them out!
  • 4! is the number of ways we can rearrange the four people consecutively within the four seats.
  • Therefore, 6 x 4! = 144 ways!
  • For this particular question, we had a really interesting debate.
  • Some said that the direction each group took was relevant but in reality, it is not!
  • Realize that we are working with combinations (where order does not matter), not a permutation (where order does matter).
  • So in the solution on the slide, we have 7 choose 4 because there are 7 people and 4 of them have two choices to take but can only take one...
  • However, we also have 3 choose 3 because the three people left only have one direction to choose from! (3 choose 3 = 1)
  • So we multiply those two together and end up with 35 ways!
  • Keep in mind that 7 choose 3 x 4 choose 4 is the same thing!

Alright, so for the poker questions, not only does Aldrin have hints for them, but John beat me to actually solving them! Here is his BOB post on Combinatorics which includes the solutions to the last few slides from today! Thanks bro!


Mmk, that's all about that comes to mind from today... Sorry if I forgot anything! Good luck on your pre-tests everyone, study hard, do homework, and don't forget to BOB!

The next scribe is... Jessi!


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