Combinations and Permutations (1 Viewer)

[allGenreGamer]

The assessment task on combination and permutations is on soon, I need help with a few final questions:

1. Letters of "promise" are arranged in a row. What's the probability that there're 3 letters between p and r? (Answer: 1/7)

2. Urn A has 6 white and 4 black balls. Urn B contains 2 white and 2 black balls. Two balls from Urn A are selected and placed in Urn B. From Urn B two balls are then selected. What's the probability that exactly 2 balls is white? (This one is HARD!, Answer: 128/225)

3. There're two blue and four yellow marbles arranged in a row. How many different arrangements of just 5 of these marbles are possible? (Answer: 25)

4. 5 travellers arrive in a town where there're 5 hotels. If two of them are husband and wife and must goto the same hotel. How many arrangements if the other three can go to any of the other hotels? (Answer: 320)

As you can see, I have the answers to these questions, but I do not understand the logic behind it. After clearing this up, I'll be confident and ready for the exam. So please come through for me again people! (you guys helped me get 97% for my 2 unit half yearlies, so be proud )

 

[allGenreGamer]

I've found some new questions that I can't do

5. There're 8 seats, 4 facing the front and 4 facing the back. Find the probability that 2 particular people will sit opposite each other if seating is arranged randomly.

6. In 12 people, 8 have seen film A and 9 have seen film B. Everyone in the group has seen at least one of these movies. If one friend is chosen, find the probability that this person has seen:
- Both movies
- Only film B

7. In how many ways can 3-letter combinations be made from the word AUSTRALIA?

Please help me... maths is just not my thing.

 

[~*HSC 4 life*~]

Originally posted by allGenreGamer
I've found some new questions that I can't do

5. There're 8 seats, 4 facing the front and 4 facing the back. Find the probability that 2 particular people will sit opposite each other if seating is arranged randomly.

6. In 12 people, 8 have seen film A and 9 have seen film B. Everyone in the group has seen at least one of these movies. If one friend is chosen, find the probability that this person has seen:
- Both movies
- Only film B

7. In how many ways can 3-letter combinations be made from the word AUSTRALIA?

Please help me... maths is just not my thing.

do you have the answers to these?

 

[nike33]

ill have a go..this topic is hard to explain...you just sort of get it...

1. Letters of "promise" are arranged in a row. What's the probability that there're 3 letters between p and r? (Answer: 1/7)

now this is 6! / 7! ..as obviously its over 7! (total arangements) now let p and r be fixed...3 spaces away from each other...therefore there are 6! ways of arranging those

umm do you want explanations for all of these or just the ones with no answers or...

 

[allGenreGamer]

Well, I think I'd understand if u just put down the working, so u dun have to write an explanation for each one. Plz continue with your assistance

~*HSC 4 life*~, I have the answers indeed:

5. There're 8 seats, 4 facing the front and 4 facing the back. Find the probability that 2 particular people will sit opposite each other if seating is arranged randomly.
(Answer: 1/7)

6. In 12 people, 8 have seen film A and 9 have seen film B. Everyone in the group has seen at least one of these movies. If one friend is chosen, find the probability that this person has seen:
- Both movies (Answer: 5/12)
- Only film B (Answer: 1/3)

7. In how many ways can 3-letter combinations be made from the word AUSTRALIA? (Answer: 60480)

 

[CM_Tutor]

I'll chip in with answers to those questions that others don't cover.

I would make the general comment that this topic is often challenging, partly because students do not really explain what they are doing. As a result, a small mistake can lead to a mark of 0 as the marker can't figure out what was done, or won't invest much time trying to understand a student's reasoning.

I like to try and describe these questions as a series of choices that must be made. Take, for example, the first question:

1. The word 'promise' has 7 different letters, with no repeated letters, and so there are 7! possible rearrangements of the letters.

We seek an arrangement with the 'p' and 'r' three letters apart. There are 6 possible arrangements of the 'p' and 'r' - p...r.., .p...r., ..p...r, r...p.., .r...p. and ..r...p
The remaining 5 letters can be placed in 5! ways.

So, there are 6 * 5! arrangements with the 'p' and 'r' 3 letters apart.

So, P(3 lletters between 'p' and 'r') = # ways with them 3 apart / # possible arrangements = 6 * 5! / 7! = 1 / 7

 

[nike33]

7. In how many ways can 3-letter combinations be made from the word AUSTRALIA? (Answer: 60480)

again excuse my poor wording/expalations..
there are 9P3 ways of getting the 3 values..

and then theres 6!/3! (accounting for the triple a) ways of arranging these.. .:. 9P3 x 6!/3!...(sum1 check this..no calculator

edit: another way..consider the full word AUSTRALIA...there are 9!/3! ways of arranging this .:. if each time this happens you split them in groups of 3..it would be the same...so 9!/3! is the answer

 

[nike33]

"6. In 12 people, 8 have seen film A and 9 have seen film B. Everyone in the group has seen at least one of these movies. If one friend is chosen, find the probability that this person has seen:
- Both movies
- Only film B"

Draw a venn diagram.. and you should be able to see that 3ppl only saw A, 5ppl saw both, 4ppl saw B....

.:. both = 5/12
only B = 4/12

if you cant see this straight away...just use simul eqns...

ie c = 12 - (a+b) (c = both, a = A only, b = B only)
a = 8 - c
b = 9 - c

adding them... a+b = 17-2c...sub that in and you get c = 5, b=4, a=3

 

[nike33]

"5. There're 8 seats, 4 facing the front and 4 facing the back. Find the probability that 2 particular people will sit opposite each other if seating is arranged randomly.
(Answer: 1/7"

let the first person sit in any seat he want...then there are 7 seats left..so there is a 1/7 chance to sit opposite to him

 

[nike33]

4. 5 travellers arrive in a town where there're 5 hotels. If two of them are husband and wife and must goto the same hotel. How many arrangements if the other three can go to any of the other hotels? (Answer: 320)

ok let the wife / hustband go into a hotel..therefore there are 4x4x4 ways of the other dudes going where thay want...but the couple can be in any of the five hotels..therefore 4x4x4x5 = 320

 

[nike33]

3. There're two blue and four yellow marbles arranged in a row. How many different arrangements of just 5 of these marbles are possible? (Answer: 25)

hmm one way...(again no calc)
6P5 / (4!x2!)

or consider the two cases. case1. one yellow is left out...5!/(2!.3!) case2. one blue is left out 5!/4! .:. answer is

5!/(2!.3!) +(5!/4!) = 15 ...uh dunno whats happenin here i gtg now anyway, ill see what i did wrong tonight

 

[allGenreGamer]

At times when I ask a question here, I doubt that anybody would bother answering me. But u guys are too much! U never dissapoint me! I need to get 95 for my UAI to enter my dream course, if I succeed, it's all thanks to you guys

 

[CM_Tutor]

Nike33 is correct on question 3 - the answer is (5! / 4!) + [5! / (3!2!)] = 15.

As far as I can see the only one left is question 2, so here is my answer, but I don't agree with the book answer:

2. Urn A has 6 white and 4 black balls. Urn B contains 2 white and 2 black balls. Two balls from Urn A are selected and placed in Urn B. From Urn B two balls are then selected. What's the probability that exactly 2 balls is white? (This one is HARD!, Answer: 128/225)

First, note that there are three possible transfers of two balls from Urn A to Urn B - two black balls, two white balls or one of each. The probabilities of each of these transfers are 2 / 15, 1 / 3 and 8 / 15, respectively.

Now, P(2 white balls from Urn B) = P(transfer was WW and select WW from 4W, 2B or transfer was BB and select WW fron 2W, 4B or trasfer was one of each and select WW from 3W, 3B)
= P(transfer was WW) * P(select WW from 4W, 2B) + P(transfer was BB) * P(select WW fron 2W, 4B) + P(trasfer was one of each) * P(select WW from 3W, 3B)
= (1 / 3) * (4 / 6) * (3 / 5) + (2 / 15) * (2 / 6) * (1 / 5) + (8 / 15) * (3 / 6) * (2 / 5)
= (2 / 15) + (2 / 225) + (8 / 75)
= (2 * 15 + 2 + 8 * 3) / 225
= 56 / 225

If someone can suggest a mistake I've made, I'd be interested - otherwise, I suggest that the book answer is wrong.

 

[nike33]

cm i got the exact thing..thats why i was reluctant to post it

 

[CM_Tutor]

Sounds like we are both right, then. Good.

 

[Xayma]

Originally posted by CM_Tutor
If someone can suggest a mistake I've made, I'd be interested - otherwise, I suggest that the book answer is wrong.

Although it isnt a problem with your results I think the question may have been typed out wrong.

As "exactly 2 balls are white" seems to indicate that there would be more then 2 balls, otherwise the exactly isnt needed.

 

[stupid idiot]

Originally posted by nike33
7. In how many ways can 3-letter combinations be made from the word AUSTRALIA? (Answer: 60480)

again excuse my poor wording/expalations..
there are 9P3 ways of getting the 3 values..

and then theres 6!/3! (accounting for the triple a) ways of arranging these.. .:. 9P3 x 6!/3!...(sum1 check this..no calculator

edit: another way..consider the full word AUSTRALIA...there are 9!/3! ways of arranging this .:. if each time this happens you split them in groups of 3..it would be the same...so 9!/3! is the answer

i do not get this at all. how come i got 42 ways for my answer?

# 3A's = 1
# 2A's = 6C1
# 1A's = 6C2
# 0A's = 6C3

# total ways = 1 + 6C1 + 6C2 + 6C3 = 42

what's wrong with my logics? did i not interpret the question right? fuck i'm stupid.

 

[zerzillia]

i hate permu/comb

 

[stupid idiot]

So no one is willing to help me?

 

[George W. Bush]

Originally posted by stupid idiot
i do not get this at all. how come i got 42 ways for my answer?

# 3A's = 1
# 2A's = 6C1
# 1A's = 6C2
# 0A's = 6C3

# total ways = 1 + 6C1 + 6C2 + 6C3 = 42

what's wrong with my logics? did i not interpret the question right? fuck i'm stupid.

Say we have no As, and we choose the letters R S and T.

According to your method, RST, RTS, TSR, TRS, SRT, STR are all the same word.