# Permutation and combination problems.

• kenny1999
In summary, the conversation is about a permutation and combination problem involving arranging 6 people, where A, B, and C must be separated from each other. The solution is 3! x 4!, which can be calculated by first arranging D, E, and F separately and then placing A, B, and C in the remaining spaces. Another approach is to place A, B, and C first and then arrange D, E, and F around them. The formula 3! x 3! + 3! x 3! + 3! x 3! + 3! x 3! also yields the correct solution of 144.
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

hi kenny!
kenny1999 said:
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)

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.

i wonder!

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

tiny-tim said:
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

well, the result is correct

but what's your justification for it?

## What is the difference between permutation and combination?

Permutation refers to the arrangement of objects in a specific order, while combination refers to the selection of objects without considering the order.

## How do I calculate the number of permutations?

The number of permutations can be calculated by using the formula n!/(n-r)! where n is the total number of objects and r is the number of objects being selected.

## What is the formula for calculating combinations?

The formula for calculating combinations is nCr = n!/r!(n-r)! where n is the total number of objects and r is the number of objects being selected.

## What is the difference between with repetition and without repetition in permutation and combination?

In permutation and combination problems with repetition, the same object can be selected multiple times, while in problems without repetition, each object can only be selected once.

## What are some real-life applications of permutation and combination?

Permutation and combination are used in various fields such as mathematics, computer science, statistics, and physics. Some real-life applications include password combinations, lottery numbers, and combination locks.

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