Probability Help PLEASE!! (2 Viewers)

[GaDaMIt]

Could someone pm me their msn so that i can bombard them with basic probability questions? This is a serious request.. i'm getting to the point where its just plain annoying me..

One such example

In how many ways can 3 maths books, 6 science books and 4 english books be placed on a shelf, if the books relating to each subject are to be kept together?

 

[rama_v]

Could someone pm me their msn so that i can bombard them with basic probability questions? This is a serious request.. i'm getting to the point where its just plain annoying me..

One such example

In how many ways can 3 maths books, 6 science books and 4 english books be placed on a shelf, if the books relating to each subject are to be kept together?

Is it
3! x (3!.6!.4!) ?

Also, don't worry too much about probability/permutations/combinations. I never understood it very well either. In the HSC, when they have a probability question, they always have an easy first part, and a hard second part which very few ppl get.

 

[GaDaMIt]

Is it
3! x (3!.6!.4!) ?

Also, don't worry too much about probability. I never understood it very well either. In the HSC, when they have a probability question, they always have an easy first part, and a hard second part which very few ppl get. So it really won't matter too much if you don't get this part

Uhh i figured it out.. its 3! x 3! x 6! x 4! which i think is what you had

im a preliminary student and we're doing the year 12 3u probability for some strange reason at our school, and i need to know this for the 3u test..

Question I am currently stuck on:

In how many ways can a boat crew of eight women be arranged if three of the women can only row on the bow side and the two others can only row on the stroke side

Edit: And the probability section of the paper will be worth approximately 30%. It's just simple probability like that except i dont get it..

So please someone pm me your msn so i can bombard you with probability questions..

 

[cutie_pie01]

Before I offer you any help, is the answer 72? because usually I do these questions a different way, but since there was no restrictions on the number of seats, I approached it differently. But if that's the answer I can show you how I did it.

 

[GaDaMIt]

Answer is 1728 ..

I have my friends book and its got working to all the questions but i don't get why he does some stuff

He had 4P3 x 4P2 x 3P3 = 1728 for that question.. i dont get why those though

 

[webby234]

Ok, so you have four people on each side, right.

Three of them must be on one side, so you are selecting three of the four ie how many ways are there to arrange three people in four seats?

4P3

Two of the four must be on the other side, so how can you arrange these two people into four seats?

4P2

The other 3 can be arranged in any order

3P3 (or 3!)

So you multiply them to get the answer.

 

[GaDaMIt]

In Morse code, letters are formed by a sequence of dashes and dots. How many different letters is it possible to represent if a maximum of ten symbols are used?

I dont even get that question ^^

 

[darkliight]

How many letters is it possible to represent, using morse code, with a maximum of one symbol?

2 right? . and -

Two symbols? Well we have 2 choices for the first symbol, a dot or a dash like before. For each one of those choices there are 2 choices for the second symbol too, which gives us 2x2 = 4 possible combinations right? (.. or -. or -- or -.).

Three symbols? 2 choices for the first symbol, 2 for the next symbol and 2 for the third symbol, i.e. 2x2x2 = 2³ = 8.

Ten symbols?

EDIT: Oops, didn't count words using less than 10 symbols.

Should be 2 + 2² + ... + 2¹⁰ = 2046. Sorry for stuffing that up.

 

[cutie_pie01]

ah, so it is 4 per side, I wasn't sure because the question didn't specify. Anyways, if you're having trouble with permutations, you could always do it with the 'box' method (as i like to call it) - I find it helps because it's more visual. I talk you through how I did it.

1) draw up 2 rows of 4 boxes (1 row for facing the bow & 1 row for facing the stern)
2) 3 women must face the bow. There are 4 seats facing the bow avaliable so you simply count down the number of seats they can choose from for each of the 3 women. Eg: woman 1 can choose from 4 seats, woman 2 can choose from 3, woman 3 can choose from 2 etc. The numbers you multiply with are the number of seats (options), not the number of people (4x3x2)
3) 2 women must face the stern. so woman 1 can choose from 4 seats, woman 2 can choose from 3 seats (4x3)
4) There are three women left over and there's 3 extra seats; so woman 1 can choose from 3 seats etc. (3x2x1)
5) Then what you do is just mulitply out the numbers
4x3x2x4x3x3x2x1 = 1728

* To make it even more visual, you could fill the numbers into the boxes as you go - as if the boxes were being taken up like the seats the boxes would look something like this (just imagine the numbers are surrounded by boxes):


4 3 2 3

4 3 2 1


then at the end you can just type all the numbers into your calculator & multiply them.

Hope it helps

 

[GaDaMIt]

Thanks alot guys.. just one more question I THINK (for the moment at least)..

5 letter words are formed without repetion from the letters of PHYSICAL

d) How many contain the letter Y?
e) How many have two vowels occurring next to one another
f) How many have the letter A immediately following the letter L

please explain each of those 3 answers ^^ =)

 

[webby234]

d) Choose the letter Y. This leaves 7 letters of which four must be chosen. The y can then be put in 5 different places so the answer is

7P4 * 5

= 4200

e) Choose I and A (the only two vowels). These are two letters of the word and will be next to each other. Then you have 6 letters of which 3 will be chosen.

6P3

The letters IA can be in 4 different places in the word (1st and 2nd, 2 and 3, 3 and 4 and 4 and 5). Furthermore you can arrange I and A in two different ways. So the answer is

6P3 * 4 * 2

= 960

f) As with e) only there is no need to multiply by two as you can not rearrange LA. So the answer is

6P3 * 4

= 480

 

[GaDaMIt]

Current Questions Requiring Help
Find how many arrangements of the letters of the word TRANSITION are possible if the Is, Ns and Ts are together?

I'm doing [5 x 6!/(2! x 2! x 2!)] x 4!.. and im getting 10800. Answer is 5400


AND


Bob is about to hang his eight shirts in the wardrobe. He has 4 different styles of shirts, 2 of each style. How many different arrangements are possible if no two identical shirts are next to one another?

 

[pLuvia]

Is the answer to the second one 16?

 

[GaDaMIt]

864..

 

[Mountain.Dew]

Current Questions Requiring Help
Find how many arrangements of the letters of the word TRANSITION are possible if the Is, Ns and Ts are together?

I'm doing [5 x 6!/(2! x 2! x 2!)] x 4!.. and im getting 10800. Answer is 5400

AND

Bob is about to hang his eight shirts in the wardrobe. He has 4 different styles of shirts, 2 of each style. How many different arrangements are possible if no two identical shirts are next to one another?

heres my 2 cents...
u do have the right expression, IE
[5! x 6!/(2! x 2! x 2!)], but you still have to halve it to get 5400 because of this: consider the letters in the 'block' I,I,N,N,T,T...we can have the permutation (INT)[INT], but that is identical to [INT](INT), where the 'INT' block switched places. Thus there is a duplicate of 'INT', not necessarily individual letters themselves.

Thus, halve your original answer to get the correct answer.

2nd question

this is my reasoning...

we have 8 shirts in total, 2 of each style. so the total number of possible arrangements = 8! / (2! * 2! * 2! * 2!) = 2520

now, the same shirts cant be together. THUS, we consider the total arrangments that the same shirts ARE together

consider two indentical shirts - AA. we have BB, CC and DD.

now, the two identical shirts can be in 7 places

_ _ _ _ _ _ _ _ <-- try placing A A side by side in the possible spaces and see for yourself.

once we put in our duplicate, it doesnt matter how we arrange the rest of the 6 clothes. AND there are 4 styles to consider as well.

therefore, the total arrangements of at least one identical shirt together is:

[4 * 7 * 6!/[2!*2!*2!]/2]= 1260

i divide by two, because of extra duplicates:

consider AABCDBCD - that has a duplicate which i havent addressed before. also consider AACBDCBD - another duplicate, etc etc...all are covered by dividing by two.

so, my final answer to the question is 1260. not sure if its right or not, but i hope the above reasoning is useful.

M.D. ^^

 

[GaDaMIt]

Let S = {1,3,5,7,9,11,13,15,17,19} be the set of the first ten positive odd integers
a) how many subsets does S have?
b) how many subsets of S contain at least 3 numbers?
c) how many subsets with at least three numbers do not contain 7?
d) how many subsets with at least three numbers do not contain 7 but do contain 19?


^^ Stuck on that now

Side question: Considering how much i dont understand.. is this normal or what?

 

[Mountain.Dew]

I got a different answer again. I thought about it this way. Lets call the shirts a1, a2, b1, b2, c1, c2, d1, and d2. Take a1, b1, c1 and d1 and arrange them anyway you like. There are 4! ways to do this. Then you are left with four shirts hanging in a row, with potential spaces between them, sort of like this

_*_*_*_*_

where the stars are the shirts and the '_' are the spaces in between. Now, take a2. We can put a2 into any space other than the two spaces next to a1. so we have 3 choices. Then take b2. This time,we have a row of 5 shirts and 6 gaps, like this

_*_*_*_*_*_

but only 4 of these gaps can be used to hang b2. Similarly, we will have 5 choices for c2 and 6 choices for d2. Now we have 4! x 3 x 4 x 5 x 6, but of course a1 and a2 are identical (as are b1 and b2, etc), so we could swap them round and still have an identical arrangement of shirts. So we must divide this number by (2!)⁴ to get the number of distinct arrangements. And when you do that, you get 540, which is not the answer that you said earlier, so I don't know what is going on...

mmmm i like your method. i think in the end, we apply this reasoning...

we HAD 5 spaces to begin with...and ended up with 8 spaces. so we have to factor that in...

final answer = 540 * 8 / 5 = 864

now, i dont know the mathematical reasoning behind it, if someone (riviet, buchanana, pLuvia, keypad...neone?) could do so it would be greatly appreciated!

NOTE: my method for the 1st attempt at this question is bunk. please ignore it.

 

[Sober]

I do not follow your reasoning, here is how I figured it, we look at all the ways that violate the conditions, those of the form of how many pairs are together:

AABBCCDD (4 pairs) = 4! = 24
AA?BB?CC (3 pairs) = 5!/2 - 24 = 36
?AA??BB? (2 pairs) = 6!/4 - 24 - 36x2 = 84
???AA??? (1 pair) = 6!/8 - 24 - 36x3 - 84x3 = 984

Then we sum each of these results by the ways they can happen using different orders of pairs = 24 + 4x36 + 6x84 + 984 = 1656, this subtracted by all the possible arrangements (8!/16 = 2520) is 864.

This probably makes absolutely no sense to anybody and is probably far more convoluted than the previous solution but I am pretty sure the question is beyond the scope of a highschool probability question.

 

[Mountain.Dew]

oooo nice one sober! although i dont fully understand how you did it, but i know how ur reasoning works.

agreed, this question is beyond high school combinatronics.

 

[GaDaMIt]

Cool.. no one seems to think im a total dumbass just yet

Anyway another question .. from my schools past 3u maths test

How many ways can the letters of the word CONNECTION be arranged if all the N's are together? Answer = 10 080 Working is 8! / (2! x 2!) .. i dont get why these numbers though.. anyone care to enlighten me?

Edit: Reason i dont get this is that the 3 Ns can be together so I'm thinking 3Ns fitting into a 10 letter slot thing being together can happen 8 ways..

So im thinking like 7! (2! x 2!) x 8

EDIT: LOL that gives the same answer.. ok so im getting it right -.- anyway.. can someone tell me why they have that working instead of my working?