# Permutation Problem

1. Sep 4, 2007

### Lemm

1. The problem statement, all variables and given/known data
In How many ways can six men and two boys be arranged in a row if:
a. The two boys are together?
b. The two boys are not together?
c. There are at least three men separating the boys?

2. Relevant equations
P= N!
(n-r)!
Identical n objects.
N!
n1 x n2 x...x nk

3. The attempt at a solution
A.Answer: 2(7!) = 10,080
B.Answer: 8!-2(7!) = 30,240

For C i have so far, divided the total into 4 groups each boy is a seperate one and 2 groups of 3 men... I set the boys as identical objects so they arent set in alternating pairs ( AB, BA would just be one) so i divide that by 2... and the 2 groups of 3 men can alternate withing each other so thats 3!^2

4! *3!*3!
2!

Im sure this is not the right way to go, and i think i have to find the total number the boys are seperated for atleast 3, 4,5,6 men and add them all.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 4, 2007

### rootX

I did this way:

ans = N-n(0)-n(1)-n(2)

so, like for n(1):

there's a group of 3 person(two boys on the ends with one man)-consider this as a unit

so there are 6 units in total, and place that big unit in one place

6C1*2 (boys interchaged their pos)

and then 6! for men..

3. Sep 4, 2007

### Lemm

I was able to find the answer on the back of the book, A and B are correct and C is 14400
I just dont get how its done.. its just a matter of picturing it i guess

4. Sep 4, 2007

### rootX

lol, I got the same!!

yea, so try picturing this way:

draw 8 boxes, and circle the first 3 (for 2 boys, and 1 man in the middle)

and consider that circle as one box

and now, you have 5 boxes, and 1 circle.

first you can place the circle anywhere (among those 6 possible positions)

and then you can place 6 men by 6! thing... (I can't explain further than this lol)

Last edited: Sep 4, 2007
5. Sep 4, 2007

### Lemm

Dont the boys have to be seperated by atleast 3 men?... whats the N-n(0)-n(1)-n(2)
is it the different scenarios... seperated by 3, 4, 5, 6, men?

Im just confused on C, i would appreciate a step by step working just to see how its done, and have it as a future reference, Thanks.

Last edited: Sep 4, 2007
6. Sep 4, 2007

### rootX

N is all possible combinations
and n(0) is when there is no one between the two boys
n(1) = 1 person is between the boys

and in there, I was finding n(1)

7. Sep 4, 2007

### Lemm

yea i didnt see ur other ''circle'' post before i qouted, yea i get it now its just i was thinking the other way like finding the 3 4 5 6 insted of getting the first, thanks

8. Sep 5, 2007

### Sleek

The answer for C is indeed 14400

As the question says atleast 3 person should be between the two boys, thus it can be 3,4,5 or 6.

Consider 3:

So lets take the two boys with three men in between as one block. This block can be placed in four different ways among the remaining three men. The boys can be interchanged (i.e. arranged among themselves in 2 ways). Also, the 6 men can be arranged in 6!

Thus, we get 4*2*6!

If there are 4 men in between, similarly above equation would become 3*2*6!, for 5 men in between, 2*2*6! and for 6 men in between, 1*2*6!.

4*2*6!=5760
3*2*6!=4320
2*2*6!=2880
1*2*6!=1440

Since at a time, only one case is possible, i.e. there can be either 3,4,5 or 6 men in between the boys and they all cannot happen simultaneously, we sum the above values.

5760+4320+2880+1440=14400

Thats a pretty good answer.

Regards,
Sleek.

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