Permutation and combination problems.

  • Thread starter kenny1999
  • Start date
  • #1
163
2

Homework Statement



I forget the exact expression of the questions. But the related details are exhaustive here.
it's about permutation and combination. By the way, I am not student, i am looking for explanation and understanding, not answers..

1. There are 6 people, namely A, B, C, D, E, F in a row, how many ways of arrangement are there such that person A,B,C must be separated

Homework Equations



No equation given


The Attempt at a Solution



I especially paid attention to the word "must be separate", then I first count the number of ways that A,B and C "must be grouped" together. which result in 3! (arrangement within A,B,C) and then multiply 4! (taken ABC as one object).

Then for 6 people randomly arranged in order there are 6!. It's easy.

so the solution i think should be 6! - 3! x 4!

however, that was wrong. the solution is 3! x 4!

I feel so hard understanding why.

I don't know what they mean by "must be separate" here, how about, for example, A E F B C D? when B and C are combined while A is separate from them. Did I forgot to minus this possibility?

The answer is actually 144 but I get no idea about it.


I am not really clever, please explain in a simple way. Thanks people. Thanks people
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
hi kenny! :wink:
I don't know what they mean by "must be separate" here, how about, for example, A E F B C D? when B and C are combined while A is separate from them. Did I forgot to minus this possibility?
yes, that's right :smile:

"separate" means no two are together

here's another way to do it:

place A B and C first, with two gaps in between (how many ways are there of doing that?)

then place D E and F (how many ways are there of doing that? you know that at least one of them has to be in the first gap)​

have another go! (both ways) :smile:
 
  • #3
132
0
Another explanation:
1. Arrange D, E, F as shown -D-E-F- , there are x ways to do that.
2. Now there are 4 positions (each denoted by a dash) to place A, B, C at.
First choose 3 positions from 4, there are y ways to do that.
Then make an arrangement among 3 peoples, this could be done in z ways.
3. Multiply things out and you'll get the answer.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
i wonder! :rolleyes:

can you calculate how many different ways are there of solving this question? :biggrin:
 
  • #5
163
2
hi kenny! :wink:


yes, that's right :smile:

"separate" means no two are together

here's another way to do it:

place A B and C first, with two gaps in between (how many ways are there of doing that?)

then place D E and F (how many ways are there of doing that? you know that at least one of them has to be in the first gap)​

have another go! (both ways) :smile:
i've worked out a formula for this problem, but is the formula correct?


3!x3! + 3!x3! + 3!x3! + 3!x3! = 144
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,832
251
well, the result is correct :smile:

but what's your justification for it? :confused:
 

Related Threads on Permutation and combination problems.

  • Last Post
Replies
2
Views
1K
Replies
2
Views
2K
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
6
Views
17K
Replies
3
Views
827
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
1K
Top