Permutations and Combinations Marathon (1 Viewer)

[Green Yoda]

To start off:
The ratio of numbers of arrangements of (2n+2) different objects taken n at a time to the number of arrangements of 2n different objects n at a time is 14:5. Find the value of n.

 

[pikachu975]

To start off:
The ratio of numbers of arrangements of (2n+2) different objects taken n at a time to the number of arrangements of 2n different objects n at a time is 14:5. Find the value of n.

(2n+2)Cn / 2nCn = 14/5
(2n+2)!/n!(2n+2-n)! * n!(2n-n)!/(2n)! = 14/5
(2n+2)!/n!(n+2)! * (n!)^2 / (2n)! = 14/5
(2n+2)(2n+1)/(n+2)(n+1) = 14/5
5(2n+2)(2n+1) = 14(n+2)(n+1)
5(4n^2 + 6n + 2) = 14(n^2 + 3n + 2)
20n^2 + 30n + 10 = 14n^2 + 42n + 28
6n^2 - 12n - 18 = 0
n^2 - 2n - 3 = 0
(n-3)(n+1) = 0
n = -1, 3 but n>0
n = 3

EDIT: It's meant to be perms not combs but still gives the same answer as dividing those 2 combinations gets rid of the n! which is the difference between the P and C.

 

[Green Yoda]

In how many ways can 5 writers and 5 artists be arranged in a circle so that two particular artists must not sit next to a particular writer. Note: The writers and artists sit alternately.

 

[bujolover]

In how many ways can 5 writers and 5 artists be arranged in a circle so that two particular artists must not sit next to a particular writer. Note: The writers and artists sit alternately.

4!*5! = 2880

 

[Green Yoda]

4!*5! = 2880

That is for when the artists are seated alternatively only. However there are other conditions to the question.

 

[pikachu975]

Fix the writer = 1
Sit 2 artists next to him = 3 x 2
Sit the rest of the writers = 4!
Sit the rest of the artists = 3!

Multiply to get 864

 

[Green Yoda]

Fix the writer = 1
Sit 2 artists next to him = 3 x 2
Sit the rest of the writers = 4!
Sit the rest of the artists = 3!

Multiply to get 864

Yup that is correct!

 

[kawaiipotato]

From the HSC 2017 MX2 Marathon thread.

(I don't have the answer unfortunately.)

 

[Green Yoda]

From the HSC 2017 MX2 Marathon thread.

(I don't have the answer unfortunately.)

I dunno if its this simple tbh but:
28C25 = 3276 ways

 

[Green Yoda]

How many ways can eight basketball players be divided into four groups of two?

 

[pikachu975]

How many ways can eight basketball players be divided into four groups of two?

8C2 x 6C2 x 4C2 x 2C2 / 4! = 28x15x6 / 4! = 105

Divided by 4! because if you pick the same teams but arrange them in different order then it will give 4! arrangements, so you gotta get rid of those as if it's a repetition when arranging EEEE etc.

 

[braintic]

A 3-player game is played between Andy, Ben and Chuck.
The probabilities that each of those players win a game are 0.5, 0.3 and 0.2 respectively.
There are no drawn games.
The winner of a tournament is the first player to win 6 games.
The current score in the tournament (ie. games won) is:
Andy - 3
Ben - 4
Chuck - 2
To 5 decimal places, what is the probability that Andy wins the tournament from here?

 

[pikachu975]

A 3-player game is played between Andy, Ben and Chuck.
The probabilities that each of those players win a game are 0.5, 0.3 and 0.2 respectively.
There are no drawn games.
The winner of a tournament is the first player to win 6 games.
The current score in the tournament (ie. games won) is:
Andy - 3
Ben - 4
Chuck - 2
To 5 decimal places, what is the probability that Andy wins the tournament from here?

Most likely completely wrong but is the answer 0.18075

 

[braintic]

Most likely completely wrong but is the answer 0.18075

Sorry, I forgot I asked this question, and I can't find my solution. I just tried it again and got 0.31275.