tag:blogger.com,1999:blog-7691251033406320222.post6500484571335058599..comments2024-03-26T04:19:38.862-07:00Comments on kitchen table math, the sequel: The New Math SAT Game PlanCatherine Johnsonhttp://www.blogger.com/profile/03347093496361370174noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-7691251033406320222.post-35893060307892617702011-03-04T12:14:27.042-08:002011-03-04T12:14:27.042-08:00My explanation would be exactly the same as NHA...My explanation would be exactly the same as NHA's. Since the question is about the office, we can safely ignore the cubicle.<br /><br />A mathematically equivalent question would be ....<br /><br />"3 members of a table tennis team have to be chosen from 3 boys and 2 girls. What is the probability of the team having exactly 1 girl?"<br /><br />(3C2 * 2C1) / 5C3 = 6/10 = 3/5<br /><br />------------------------<br /><br />After having said that, I also like to check the same answer with basic probability .... slot filling, if you will. If I can't do a question both ways and get the same answer ..... :-(<br /><br />1st slot = boy = 3/5<br />2nd slot = boy = 2/4 <br />3rd slot = girl = 2/3<br /><br />There are three possible slots that the girl could be chosen for so ..<br /><br />(3/5 * 2/4 * 2/3) * 3 = 3/5<br /><br /><br />Richard IAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-17618649567186470812011-03-03T07:18:24.097-08:002011-03-03T07:18:24.097-08:00http://www.coolmath.com/algebra/20-combinatorics/i...http://www.coolmath.com/algebra/20-combinatorics/index.html<br /><br />also a straightforward free concept explanationlgmnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-228411362202533932011-03-02T19:11:31.958-08:002011-03-02T19:11:31.958-08:00Regarding resources on this topic, I am finding Ar...Regarding resources on this topic, I am finding Art of Problem Solving's book "Competition Math for Middle School" to be very helpful. It is straightforward, has many worked examples, and the problem sets have solutions explaining how to answer each problem.<br /><br />-KarenAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-51581616833581061492011-03-02T18:55:52.215-08:002011-03-02T18:55:52.215-08:00Catherine,
The basic counting rule comes into pl...Catherine, <br /><br />The basic counting rule comes into play this way:<br /><br />Sometimes, instead of thinking "slots to fill," it's easier to think of "tasks to complete." In particular, that's how we want to think of this problem you've stated:<br /><br />Task 1: Choose the two men (order unimportant) <br />Task 2: Choose the woman (order unimportant)<br /><br />Basic counting rule: <br />(# of ways of performing task 1) X (# of ways of performing task 2)<br /><br />Thus, we have (# of ways to choose 2 of 3 men) X (# of ways to choose 1 of 2 women), which is (using notation) C(3,2) X C(2,1) = 3 X 2 = 6<br /><br />This takes care of the numerator of your probability fraction. The denominator, of course, is the total number of ways to choose 3 out of the 5 for the offices, without regard to either order or gender. Thus, the denominator of the probability fraction is C(5,3) = 5!/(3! X 2!) = 10.<br /><br />Hence: probability of stated problem is 6/10 = 3/5.<br /><br />Hope I've got that right ;) I will readily admit that counting problems can seem tricky if you're not accustomed to doing them. Every semester when I come to this section, I have to re-read the counting stuff to make sure that I remember enough of the details to be able to do these problems more or less off the cuff.Niels Henrik Abelhttps://www.blogger.com/profile/00554447042962336254noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-37100446334721114172011-03-02T14:34:34.206-08:002011-03-02T14:34:34.206-08:00Oops, please delete that last paragraph -- it was ...Oops, please delete that last paragraph -- it was left over from a first draft and didn't get edited out. (Although it allows you to see my tracks: My first time through I focused on filling the offices, not the cubicles, which turns out to be a more complicated way of thinking about it. In particular I was mistaken in saying the overcount was a factor of 3*2, when it's actually just a factor of 2.)Michael Weissnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-72138011953443705462011-03-02T14:32:17.208-08:002011-03-02T14:32:17.208-08:00(btw, my first comment is supposed to lead you to ...(btw, my first comment is supposed to lead you to answer why the "how many choices I have left" gave you 6, and why that was wrong...)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-47049224193682409682011-03-02T14:32:14.459-08:002011-03-02T14:32:14.459-08:00Let's do this in two parts.
Part 1 is: How ma...Let's do this in two parts.<br />Part 1 is: How many different ways of assigning people to the various workspaces are there, in total?<br />Part 2 is: How many different ways are there that meet the conditions specified by the problem?<br /><br />Then we'll divide the result of Part 2 by the result of Part 1, to get a probability.<br /><br />So, Part 1:<br />We need to choose 2 people who get cubicles. We have 5 choices for the first cubicle; then 4 choices for the second one. Once those choices are made, there are no more choices; the remaining 3 people all get offices. So in all there are 5*4 decisions to make. However, this is actually an overcount -- it treats "Jim in cubicle 1 and Paul in cubicle 2" as a different configuration from "Paul in cubicle 2 and Jim in cubicle 1". So we need to divide that result by 2. Therefore there are 10 ways of assigning people to workspaces.<br /><br />Now for Part 2. We need to choose a <i>man</i> for one cubicle and a <i>woman</i> for the other. There are 3 choices for the man who gets a cubicle; then there are 2 choices for the woman who gets a cubicle. All together there are 3*2 choices. This time there is no overcount, because if "Bob in the male cubicle and Nancy in the female cubicle" is one configuration, you do NOT get another by switching the names around. So there are 6 configurations.<br /><br />Synthesis: In 60% of the arrangements, the cubicles end up assigned to 1 man and 1 woman, with the offices going to the other 2 men and the other 1 woman.<br /><br />Now for Part 2. We want to choose a man to the first office; there are 3 choices. Having done that, we want to choose another man for the second office; now there are only 2 choices. Having done that, we assign a woman to the last office: there are 2 choices. All together we have 3*2*2 choices. Again, this is an overcount; we need to divide by the number of rearrangements of 3 people, which (as before) is 3*2. So all together there are (3*2*2)/(3*2) ways of assigning the offices in accordance with the specified constraints. Having done so, the remaining man and woman get the cubicles automatically, so no more choices get made. There are thus 2 ways of doing this. (In hindsight, those 2 choices amount toMichael Weissnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-25029420577345384642011-03-02T14:31:10.030-08:002011-03-02T14:31:10.030-08:00(read this comment later, after you've thought...(read this comment later, after you've thought for a while about the above) <br /><br />But let me add that you're supposed to be proficient enough with counting by this problem that you "See" another "trick", called the pigeonhole principle.<br /><br />What if the problem had asked how many ways are there to assign 2 women to the cubicle? Or how many ways are there to assign two women and 1 man to the office? would you have started counting out the ways to deal with all 5 employees?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-67361026815652638912011-03-02T14:25:32.607-08:002011-03-02T14:25:32.607-08:00I'm not yet answering your question, but tryin...I'm not yet answering your question, but trying to slow you down and start answering pieces of this problem first.<br /><br />when people are assigned ways of seating in seats, then ORDER matters. There's a clear notion someone goes on the leftmost seat, someone goes on the next leftmost seat that remains, etc.<br /><br />So, the assignment in order of<br />Catherine, Ed, C, A, J, and S<br />is different than <br />Ed Catherine C A J S<br /><br />When you assign people to a room, order doesn't matter.<br /><br />Are there still 60 ways to put 3 people in one room and 2 in another?<br /><br />Saying "I chose Catherine, that leaves 4 choices for the second slot, I pick Ed, that leaves 3, I choose A"<br /><br />you'll count "I chose Ed, that leaves 4 choices, I pick Catherine, that leaves 3, I choose A..." as two of your 60.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-53642327044022709152011-03-02T13:08:13.786-08:002011-03-02T13:08:13.786-08:00Hi Niels!
Now I'm having trouble figuring out...Hi Niels!<br /><br />Now I'm having trouble figuring out combinations .... at least, I think that's what I'm having trouble with.<br /><br />I can only easily solve the first part of this SAT problem:<br /><br /><i>Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?</i><br /><br />I get that there are 60 ways to choose 3 employees for the office.<br /><br />But that's all.<br /><br />I particularly don't understand this part of the solution:<br /><br /><i>From the 3 male workers, 2 can be chosen in 3 different ways.</i><br /><br />I understand it when I write out the possibilities, but I don't understand it when I try to apply the "How many choices do I have?" approach.<br /><br />When I ask myself how many choices I have left, I end up with 3x2.Catherine Johnsonhttps://www.blogger.com/profile/03347093496361370174noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-25810655942555462222011-03-02T12:53:34.987-08:002011-03-02T12:53:34.987-08:00That is exactly what I do with MATHCOUNTS problems...That is exactly what I do with MATHCOUNTS problems. I can never remember the formulas and even if I could, the problems always have some twist that makes them useless. <br /><br />But when you walk through everything and ask how many choices do I have now (you also need to ask whether order matters -- if not you need to divide by the number of ways to arrange things to eliminate double counting) it works much better.Daniel Ethierhttps://www.blogger.com/profile/11009564524804093995noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-30056854131352534652011-03-02T11:58:46.078-08:002011-03-02T11:58:46.078-08:00That's pretty much how I approach it when I ha...That's pretty much how I approach it when I have to explain the basic counting rule: <br /><br />How many "slots" do you have to fill? (Draw the slots on the board.) <br /><br />How many choices are there for each slot? (Write the appropriate number in each slot.)<br /><br />Put a multiplication sign in between each number, and an equal sign at the end. Multiply.<br /><br />Permutations of n distinct objects taken r at a time can also be treated in this manner - e.g., if there are 100 participants in a contest, in how many ways can 1st, 2nd, and 3rd place prizes be awarded?Niels Henrik Abelhttps://www.blogger.com/profile/00554447042962336254noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-7998638484576742482011-03-02T10:52:33.366-08:002011-03-02T10:52:33.366-08:00OK, I spoke too soon.
Still a great book, though...OK, I spoke too soon.<br /><br /><br />Still a great book, though.Catherine Johnsonhttps://www.blogger.com/profile/03347093496361370174noreply@blogger.com