SIGH! Probability Question. (1 Viewer)

[whackymac]

Six men and five women are arranged at random in a row so that each woman is between two men.

i) how many such arrangements are possible?

ii) what is the probability that two specified men, C and D, sit at the ends of the row?


PLEASES show working out, i want to improve on this miserable topic!

 

[EvoRevolution]

(1)


ok iam not sure if it is write but that is my go, someone check if it is write, it looks logical

(2)
2x4x3x2x1x5x4x3x2x1x1
=5760
opps probabilty soz

p() = 5760/5!6!
= 1/15
again iam not sure if it is write someone check

 

[Al-Ghazali]

6! x 5!

M W M W M W M W M W M

the men can be arranged in 6! ways
women can be arranged 5! ways


B)


C W M W M W M W M W D

C and D can be arranged 2! ways
remaining men can be arranged 4! ways
women can be arranged 5! ways


= 2! x 4! x 5! ways

probability = (2! x 4! x 5!) / (5!6!)

 

[Timothy.Siu]

Six men and five women are arranged at random in a row so that each woman is between two men.

i) how many such arrangements are possible?

ii) what is the probability that two specified men, C and D, sit at the ends of the row?


PLEASES show working out, i want to improve on this miserable topic!

i) 6!x7P5
ii)(2x4!x5!)/(6!x7P5)

 

[EvoRevolution]

woo tim what did u do

 

[Timothy.Siu]

your solution doesn't include the cases like WMMWMWMWMWM when theres theres 2 men next to each other inbetween the women

 

[Aquawhite]

your solution doesn't include the cases like WMMWMWMWMWM when theres theres 2 men next to each other inbetween the women

Ah, clever... I probably wouldn't have thought of that under exam conditions...

 

[FFC]

your solution doesn't include the cases like WMMWMWMWMWM when theres theres 2 men next to each other inbetween the women

That situation is not important, as the women must be "in between TWO men"

 

[Timothy.Siu]

That situation is not important, as the women must be "in between TWO men"

yeah your're right