Permutations Questions (2 Viewers)

[Twickel]

Hi
In how many ways can 6 people be seated in a motor car if only 2 can occupy the drivers position?

For the amount of ways we can arrange all 6 of them in any place = 6!

Now pretend the people are ABCDEF
we can have AB BA CD DC EF FE driving but AB= BA, Cd=DC Ef=FE so three possibilites is the answer 720/3=240

Well the answer is 240, I just need to know if wmy method was correct

Thanks

 

[YannY]

Hi
In how many ways can 6 people be seated in a motor car if only 2 can occupy the drivers position?

For the amount of ways we can arrange all 6 of them in any place = 6!

Now pretend the people are ABCDEF
we can have AB BA CD DC EF FE driving but AB= BA, Cd=DC Ef=FE so three possibilites is the answer 720/3=240

Well the answer is 240, I just need to know if wmy method was correct

Thanks

Well you say AB=BA then this is not a permutations question, it then becomes combinations

 

[YannY]

I think this is how you should do it:

Regard the front drivers as one person then figure out the possibility then times it by two since the front drivers can have two possibilites when everyone else is seated. i.e 5p5x2=240ways.

 

[tacogym27101990]

i reckon yanny's ways the right way id do it 2 X 5!=240
as there are only 2 ppl that can sit in the front drivers and when one person is sitting in the front seat, theres 5! ways the other ppl can sit

 

[aakash]

hmmm...Yanny's way looks right. The only doubt i have- is the car a six-seater or the usual one which has 5 seats. but interestingly that wont change the answer...
If u do assuming there are 5 seats then: no. of ways to choose the driver = 2
no. of ways to choose and arrange 4 out of 5 remaining = 5p4
so total ways = 2 X 5p4 = 240
Just a suggestion - if u get this type of question in exam and you're confused about what the question in exactly asking then write what you assumed is was (like here i said-"assuming there are 5 seats").
You would get most of the marks unless you reduced the hardness of the question due to your assumption.

Now pretend the people are ABCDEF
we can have AB BA CD DC EF FE driving but AB= BA, Cd=DC Ef=FE so three possibilites is the answer 720/3=240

Well the answer is 240, I just need to know if wmy method was correct

looks like you are forcing the answer to be 240...
a better explanation would be- only 1/3 of the total people can occupy drivers seat and hence the answer would be 1/3 of total ways.
This works in this case but i suggest not to use this.

Another way could be-
6! - 4* 5p5
6! = no. of ways to arrange all 6 of them
4* 5p4 = removing the cases when A and B are not drivers

 

[YannY]

I look at this from a lines perspective. Only two person can stand in front of the line.

Therefore to fill the front there are two ways.

To fill the second position there are 5 people left.

To fill the 3rd position there are 4 people left.

goes on to the last position where thereis only one person left.

hence its 2x5x4x3x2x1.

If its a five seater then that one person is exluded from the queue but that wont make a difference as he cannot change the number of ways they arrange but the method is

2x5x4x3x2

 

[aakash]

I look at this from a lines perspective. Only two person can stand in front of the line.

Therefore to fill the front there are two ways.

To fill the second position there are 5 people left.

To fill the 3rd position there are 4 people left.

goes on to the last position where thereis only one person left.

hence its 2x5x4x3x2x1.

If its a five seater then that one person is exluded from the queue but that wont make a difference as he cannot change the number of ways they arrange but the method is

2x5x4x3x2

Yep thats good enough