Permutations and Combinations.... HELP (1 Viewer)

[SnowyAngel]

SUPER stuck with perms and combs, help me!!
1. Five travellers arrive in a town where there are five hotels.
i) how many different accommodation arrangements are there if there are no restrictions on where the travellers stay?
ii) how many different accommodation arrangements are there if each traveller stays at a different hotel?
iii) suppose two of the travellers are husband and wife and must go to the same hotel. How many different accomodation arrangements are tehre if the other three can go to any of the other hotels?

2. Katie is one of ten members of a social club. Each week one member is selected at random to win a prize.
i) what is the probability that in the first 7 weeks Katie will win at least 1 prize?
ii) show that in the first 20 weeks Katie has a greater chance of winning exactly 2 prizes than of winning exactly 1 prize.
iii) for how many weeks must katie participate in the prize drawing so that she has a greater chance of winning exactly 3 prizes than of winning exactly 2 prizes?

3. A tank contains 10 tagged fish and 50 untagged fish. On each day, 4 fish are selected at random from the tank and placed together in a separate tank for observation. Later the same day the 4 fish are returned to the original tank.
i) what is the probability of selecting no tagged fish on a given day?
ii) what is the probability of selecting at least one tagged fish on a given day?
iii) calculate the probability of selecting no tagged fish on everyday for 7 given days.
iv) wat is the probability of selecting no tagged fish on exactly 2 of the 7 days.

 

[Sy123]

Are the answers to the first one:
i) 550
ii)120
iii)300

Just making sure before I start explaining.

 

[Aykay]

1: i) Each traveller can stay in each hotel, so 5^5 = 3125
ii) 5 can stay in the first one (any one) then there are 4 remaining that can stay in the second, then 3, then 2, then 1. So its 5! = 120
iii) You group the husband and wife as one set. So say there are 4 'individual' items now, since the husband and wife are being counted as 1. The husband and wife can choose any 1 of the 5 hotels. The remaining 3 can choose any of the other 4, so its 5 * 4^3 = 320.

Are the answers to the first one:
i) 545
ii)120
iii)300

Just making sure before I start explaining.

Not sure what you're doing..

 

[Sy123]

Are the answers to the first one:
i) 545
ii)120
iii)300

Just making sure before I start explaining.

i) We need to group our travellers in accordance to which hotels they go in, if you go into hotel A you are preson A, if hotel B, person B and so on.
So we need to find how many ways we can arrange:

ABCDE

AACDE

AAADE

AAAAE

AAAAA

But remember each of these arrangements can be made with any of the 5 letters, hence we have to times 5 to each event (except for the first one):

So number of arrangements are:


=550 (different from my previous answer I know I didnt count AAAAA)

ii)Normal Arrangements 1 at a time = 5! =120

iii)Take arrangements 2 at a time = \frac{5!}{2!} \times 5 = 300

I dont know what could be wrong with my logic

 

[Aykay]

i) We need to group our travellers in accordance to which hotels they go in, if you go into hotel A you are preson A, if hotel B, person B and so on.
So we need to find how many ways we can arrange:

ABCDE

AACDE

AAADE

AAAAE

AAAAA

But remember each of these arrangements can be made with any of the 5 letters, hence we have to times 5 to each event (except for the first one):

So number of arrangements are:


=550 (different from my previous answer I know I didnt count AAAAA)

ii)Normal Arrangements 1 at a time = 5! =120

iii)Take arrangements 2 at a time = \frac{5!}{2!} \times 5 = 300

I dont know what could be wrong with my logic

I think you may be misinterpreting the question.

For i):

There are 5 people, and 5 hotels.
So in each hotel, ANY of the 5 people can stay at them, since there are no restrictions.
So it's 5 x 5 x 5 x 5 x 5 = 5! = 3125.

I'm not to sure why you're using cases (addition).

ii) We both agree on that.

iii) Okay, say the 5 travelers are A B C D and E.

A and B can be the couple, but they always have to stay in the same hotel. So you can group A and B and now your "4" travelers are AB C D and E.

AB can go in any of the 5 hotels. C D and E can go in any of the remaining 4, and there are no restrictions on them, so like the first question, any 3 can stay in any of the 4 hotels left, so you have 4 x 4 x 4 = 4^3
Multiplying both parts, you get 5 x 4^3 = 320.

 

[RivalryofTroll]

I think you may be misinterpreting the question.

For i):

There are 5 people, and 5 hotels.
So in each hotel, ANY of the 5 people can stay at them, since there are no restrictions.
So it's 5 x 5 x 5 x 5 x 5 = 5! = 3125.

I'm not to sure why you're using cases (addition).

ii) We both agree on that.

iii) Okay, say the 5 travelers are A B C D and E.

A and B can be the couple, but they always have to stay in the same hotel. So you can group A and B and now your "4" travelers are AB C D and E.

AB can go in any of the 5 hotels. C D and E can go in any of the remaining 4, and there are no restrictions on them, so like the first question, any 3 can stay in any of the 4 hotels left, so you have 4 x 4 x 4 = 4^3
Multiplying both parts, you get 5 x 4^3 = 320.

This is correct.

 

[Sy123]

I think you may be misinterpreting the question.

For i):

There are 5 people, and 5 hotels.
So in each hotel, ANY of the 5 people can stay at them, since there are no restrictions.
So it's 5 x 5 x 5 x 5 x 5 = 5! = 3125.

I'm not to sure why you're using cases (addition).

ii) We both agree on that.

iii) Okay, say the 5 travelers are A B C D and E.

A and B can be the couple, but they always have to stay in the same hotel. So you can group A and B and now your "4" travelers are AB C D and E.

AB can go in any of the 5 hotels. C D and E can go in any of the remaining 4, and there are no restrictions on them, so like the first question, any 3 can stay in any of the 4 hotels left, so you have 4 x 4 x 4 = 4^3
Multiplying both parts, you get 5 x 4^3 = 320.

Ok I get it, thanks man