Problem Solving

first off, note that these forums format text in a way such that the widths of different characters are NOT constant. what this means for you, among other things, is that you can't write vertical fractions (as you've attempted to do in this post), as they'll come out badly deformed.

therefore, if you're going to write fractions involving long expressions, write them in the following way:
(10!) / (3! x 7!)
such fractions won't be distorted by the way text appears in the forum posts.

Enginpasa1 wrote: Specificaly, I dont know wen to divide by what?

here's the basic idea:

if you have a set of things that are interchangeable - meaning that, if you switch them around, the situation is the same as it was before the switch - then this is an 'order doesn't matter' type of situation. so, in this sort of situation, you need to divide by the factorial of the number of interchangeable items.
if you're using our anagram method, then these sorts of items get the same letter in the anagram.

if you have a set of things that are not interchangeable, meaning that the situation is different if any of them are switched or permuted, then 'order matters', and you don't divide by the factorial in the fractional expression.
if you're using our anagram method, then these sorts of items get different letters in the anagram.

Enginpasa1 wrote:ex. 1) The principal pf a high school needs to schedule observations of 6 teachers. SH eplans to visist one teacher each day for a week, so she will only have time to see 5 of the teachers. How many different observation schedules can she create?

my approach: 6x5x4x3x2x1 6!
----------------- = ------------------ 6 but 6 teachers so 36
y n n n n n 5!


qa is 720 and not dividing by anything.

here's the deal with this one: the principal sees each teacher on a different day, mon-fri. what this means is that the five observation slots are NOT interchangeable. specifically, the observation schedule Smith, Jones, Brown, White, Williams is different from the observation schedule Jones, Smith, Brown, White, Williams: in the first schedule Smith is seen on Monday and Jones on Tuesday, but in the second Smith is seen on Tuesday and Jones on Monday.

therefore, you can't divide by anything, as each of the positions in which a teacher can be placed (mon, tues, wed, thurs, fri, none) is distinct from any other position. therefore, nothing is interchangeable with anything else.

if you're going to do this problem with an anagram, the anagram should be
MTWHFN (where H stands for thursday)
instead of YYYYYn.

--

if the crux of the problem were merely to select which five of the six teachers would be reviewed - without placing them on specific days in the review schedule - then your answer would be correct.

Enginpasa1 wrote:ex.2) My MAIN PROBLEm-
The New York Classical Group is designing the liner notes for an upcoming cd release. THere are 10 soloists featured on the album, but the liner notes are only 5 pages long, abd therefore only have rom for 5 soloists. The solosits are fighting over which of them will appear in the liner notes, as well as who whill be featured on which page. How many different liner note arrangements are abailable?

my approach: 10x9x8x7x6x5x4x3x2x1 10! 10!
---------------------------- = --------- or --------------
y y y y y n n n n n 5! 5! 5!


I am not sure on which one is the correct way.
lost on where to divide by and by what?

again, follow the rule stated above: things that are interchangeable get the same letters in the anagram, while things that aren't don't.

in this case:
* the five musicians who get into the liner notes are NOT interchangeable, because they are placed on pages 1, 2, 3, 4, 5 in some specified order.
* the five musicians who don't get into the notes are interchangeable, because you can switch two 'no's without changing anything.

therefore, the correct anagram is
12345NNNNN
and the correct fraction is
(10!) / (5!)

notice that this problem can be solved by successive choice (the 'line method' or 'slot method'): you just need to pick who goes on page one, who goes on page two, etc. there are 10 choices for page one, 9 remaining choices for page two, and so forth, so the total number of choices is 10 x 9 x 8 x 7 x 6. this looks different, but it's the same as (10!)/(5!), as you can see by expanding the factorials and cancelling the common terms.

Enginpasa1 wrote:Also as a side note- when applying some simple logic and have two different cases and must add the two situations; we multiply when situations are joined by AND and add when situations are joined by or..correct me if I am wrong?[/quote[

that's a pretty good statement of the rule, although it's important to note that the 'and' and 'or' are not parallel. specifically:
- the 'and' you're talking about means that more than one choice is being made. for instance, the question above, about liner notes, represents five different decisions: who's on page one, who's on page two, and so on.
- the 'or', as used in official problems, generally applies to alternative options for ONE choice. therefore, while the test will commonly ask something like 'what is the # of outcomes such that the coin comes up heads once or twice?' - two distinct outcomes for the same decision/event - you will rarely, if ever, see something like 'how many committees can be selected so that the president is male or the vice-president is female?' - two different decisions. in the obnoxious event that you get the latter type of question, you can't add the probabilities anyway; you'd have to use the A + B - AandB formula that also applies to probability problems.

Enginpasa1 wrote:3.) A second grade class is writing reports on birds. The students teacher has given them a list of 6 birds they can choose to write about. If lizzie wants to write a report that includes two or three of the birds, how many different reports can she write?

3 birds situation: 6x5x4x3x2x1 6!
------------------ = ----------
N N N Y Y Y 3!

2 birds situation: 6x5x4x3x2x1 6!
------------------= ----------
N N Y Y Y Y 2!


I was wrong but dont see how to decipher the reasoning why soluton is better. solutoion is 30 when I got 150

you are correct that you need to add the # of possible reports with two birds to the # of possible reports with three birds.

also, you set up the anagrams in a COMPLETELY correct fashion, assuming that you allow the strange convention that 'N' stands for 'included' and 'Y' stands for 'not included' (the only way to justify the four Y's and two N's for the reports with two birds).

so ... you need to follow through on your own anagrams!
both the Y's and the N's are indistinguishable, so you need to divide by both factorials. so:
the first fraction should be (6!) / (3!3!) = 20;
the second fraction should be (6!) / (2!4! = 15.
the solution should therefore be 35.

not sure where the 30 comes from. that would be the correct answer if the problem were 'either 2 or 4 birds', but it isn't.

freedsl wrote:is there a link to the "anagram method" that details it.

the anagram method forms a large part of the combinatorics chapter in manhattangmat's word translations strategy guide.

as far as i know, there's no 'chapter' online that explains the anagram grid, so you'll probably have to get a hold of the book for the full explanation.

i can give you a brief rundown:

the anagram grid represents the different groups / spots / etc. to which people or items are being assigned as letters or numbers in an anagram. for instance, if we're choosing a starting lineup of 2 guards, 2 forwards, and 1 center from a basketball team with 12 players, then the twelve players would get the following letters: two G's (for the guards), two F's (for the forwards), one C (for the center), and seven N's (for the seven players who didn't make it into the starting lineup).

these anagrams could then be arranged in 12! / (2!2!7!) different ways: note that you have to divide by the factorial of each of the indistinguishable groups (here, you can't tell the 2 guard positions apart, can't tell the 2 forward positions apart, and of course can't tell the 7 non-starting positions apart).

that's a brief rundown. if you want more details, either search this forum for the term 'anagram grid' (using the search box at the corner of the page) or go grab the book.

good luck!