# Qr

#### prsndwg

7+ Year Member
If 4 ppl are arranged at random in a line, what is the probabilty of 2 particular being next to eachother??

Do you use premutation or combination rule?

Thanks

#### UCfan

10+ Year Member
If 4 ppl are arranged at random in a line, what is the probabilty of 2 particular being next to eachother??

Do you use premutation or combination rule?

Thanks
4!/2!=(4*3*2*1)/(2*1)=12. Permutation.

Proof (assume 4 people A,B,C,D)

A,B,C,D
A,B,D,C
C,A,B,D
D,A,B,C
C,D,A,B
D,C,A,B

(the order of A and B is not an issue- problem statement)

A and B can switch that make it 12.

#### dentrilla

10+ Year Member
4!/2!=(4*3*2*1)/(2*1)=12. Permutation.

Proof (assume 4 people A,B,C,D)

A,B,C,D
A,B,D,C
C,A,B,D
D,A,B,C
C,D,A,B
D,C,A,B

(the order of A and B is not an issue- problem statement)

A and B can switch that make it 12.

the probabitily would be 12/4! = 1/2... correct me if im wrong I came to this a different way... The chance that ur buddy will sit beside u is (1/4) chance he sits on one side of you + (1/4) he sits on the other side.. giving you (1/4)+(1/4) = 1/2

Not sure if this way of thinking is correct tho, someone lemme know.. THanks!

#### anteater85

7+ Year Member
If 4 ppl are arranged at random in a line, what is the probabilty of 2 particular being next to eachother??

Do you use premutation or combination rule?

Thanks

Hi,
since the order matters for this question use this formula:
N!/ (N-k)!

your N is the total number of people which is 4
your K is the desired or particular order that you want to get

so 4!/(2!) =12

#### dentrilla

10+ Year Member
Hi,
since the order matters for this question use this formula:
N!/ (N-k)!

your N is the total number of people which is 4
your K is the desired or particular order that you want to get

so 4!/(2!) =12
12 is not a probability -_-

#### dentrilla

10+ Year Member
Hi,
since the order matters for this question use this formula:
N!/ (N-k)!

your N is the total number of people which is 4
your K is the desired or particular order that you want to get

so 4!/(2!) =12

o yah and order does NOT matter. A can be infront of B or B can be infront of A. When you talk about order you must look at if the order of the group that your selecting and see if that order matters. if this case they just need to be beside eachother, not in order.

#### prsndwg

7+ Year Member
o yah and order does NOT matter. A can be infront of B or B can be infront of A. When you talk about order you must look at if the order of the group that your selecting and see if that order matters. if this case they just need to be beside eachother, not in order.

So if the order doesnt matter, then would you use recombination rule?

#### dentrilla

10+ Year Member
So if the order doesnt matter, then would you use recombination rule?

Hmm not sure what you mean by that. Remember when order doesnt matter your gonna have more options then when order does matter. So you use the permutation to find out how many options there are for the two people to be together. (12). After that you look at how many total ways 4 people can be rearranged, which is 4! = 24. so if there are 24 ways 4 people can be arranged and 12 of those ways are with you and ur bud sitting beside eachother then you have 50% chance of sitting beside your bud!

#### UCfan

10+ Year Member
So if the order doesnt matter, then would you use recombination rule?
Yes the answer needs to be 1/2. If you see there are 12 ways to arrange two of them right next to each other. All of them can be order 4!=4*3*2*1=24

So

12/24=1/2

#### prsndwg

7+ Year Member
Yes the answer needs to be 1/2. If you see there are 12 ways to arrange two of them right next to each other. All of them can be order 4!=4*3*2*1=24

So

12/24=1/2
so there isnt a one formula for the whole thing ha?

#### UCfan

10+ Year Member
so there isnt a one formula for the whole thing ha?
At first your approach is wrong (1/4+1/4). You did not consider the number of people in the group. Just think that 5 people are in the group and you want two of them next to each other. The number of sequence 5!/3!=5*4=20. The total is 5!=120. The probability 20/120=1/6. It is not equal to 1/5+1/5=2/5.
I do not know any other formula to get an answer in one step!

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#### dentrilla

10+ Year Member
At first your approach is wrong (1/4+1/4). You did not consider the number of people in the group. Just think that 5 people are in the group and you want two of them next to each other. The number of sequence 5!/2!=5*4=20. The total is 5!=120. The probability 20/120=1/6. It is not equal to 1/5+1/5=2/5.
I do not know any other formula to get an answer in one step!
5!/2! = 5*4*3 = 60

but yah about the 1/4+1/4 thing i was just seeing if that train of thought is correct ( i know how to do it the normal way) but yah when u mix up the numbers i dont think its valid.

#### UCfan

10+ Year Member
5!/2! = 5*4*3 = 60

but yah about the 1/4+1/4 thing i was just seeing if that train of thought is correct ( i know how to do it the normal way) but yah when u mix up the numbers i dont think its valid.
Sorry I made typo mistake! 5!/3! (5!/(5-2)!) not 5!/2!. That s not 60 (20).1/2 is really high probability!