Permutation and combination problems.

[kenny1999]

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

 

[tiny-tim]

hi kenny!

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

"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)

 

[drawar]

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.

 

[tiny-tim]

i wonder! …

can you calculate how many different ways are there of solving this question? ​

 

[kenny1999]

hi kenny!


yes, that's right

"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)

i've worked out a formula for this problem, but is the formula correct?


3!x3! + 3!x3! + 3!x3! + 3!x3! = 144

 

[tiny-tim]

well, the result is correct

but what's your justification for it? ​