permutation or combination?

Question

when the 3 children are go to school they each took one of the others' lunches. in how many ways could the 3 children have been paired with the luches so that on one took his or her own?

Is it permution ? how to solve those kind of question?????

no no the answer is 2 ways but I don't know how

I don't want that answer I want a formula such as (3x2x1) /3 =2

Oh never mind I solved it already! what's happening to me,

I asked now I answered it myself O0O

Answer 1

Combination is where order does not matter.
Permutation is where order matters.

For example, if I have 5 numbers, 1 through 5, permutations of this set could be:

1, 3, 4, 5, 2
2, 4, 1, 3, 5
etc.

I could also have 3 number permutations of this set:
1, 2, 3
1, 3, 2
1, 4, 5
4, 1, 5
etc.

In general, the number of r-sized permutations of a size n set is:

n! / (n-r)!

Where n! = n * (n-1) * ... * 2 * 1. 0! = 0

In contrast, a combination disregards order. If I have 5 numbers, and I want to chose three of them, I could choose:

1, 2, 3
1, 2, 4
1, 2, 5
etc.

(Note that 1, 2, 3 and 1, 3, 2 are the same.)

In general, the number of k-sized combinations in a size n set is:

n! / k!(n-k)!

In your problem, there are 3 lunches, and 3 kids.
Therefore, since ordering matters, it is a permutation (order matters).

3! / (3-3)! = 6/1 = 6

Thus, there are 6 ways to order the lunches.

If I read your typo correctly, it means that nobody can have their own lunch.

Let's enumerate the cases:

Kid A Kid B Kid C
--------------------------
A B C
A C B
B A C
B C A
C B A
C A B

We need to eliminate all the cases where any kid has their own lunch.

Eliminate the following cases:
Kid A Kid B Kid C
--------------------------
A B C <==
A C B <==
B A C <==
B C A
C B A <==
C A B

Thus, the answer is 2.

Answer 2

Suppose the lucnhes are numbered 1, 2, 3, with each lunch belonging to the corresponding child. So child 1 has lunch 1, child 2-lunch 2, and child 3- lunch 3. How many options for them to choose lunches, but not select their own. We can start the examination with child 1. He cannot have lunch 1, which is his own, so he has the two options: lunch 2 or lunch 3. If he selects lunch 2, let us look at child 3. Child 3 now cannot select his own, which means lunch 3 is excluded. But, because of the previous choice of lunch 2, this one is also gone. So his only option is lunch 1. Then child number 2 is left with the only remaining lunch, number 3.
In this scenario child 1 has lunch 2, child 3 selects lunch 1 and finally child 2 selects lunch number 3. This is a viable scenario, because none of children selected their lunches, so it satisfies the conditions. This scenario was based on the choice of child 1, but he had two options for his selection. He couldn't select lunch 1, which was his own, and we already saw what hapens when he selects lunch 2. So now we will examine the remaining choice, of lunch 3. Suppose then that child 1 selects lunch 3. Let us look this time at child 2. He cannot select lunch 2, which is his own, nor can he select lunch 3, previously chosen by child 1. So his only choice is to have lunch 1. The final child, number 3 cannot select lunch 1 or lunch 3, already chosen by the other two children. So he can only choose lunch 2. That means that child 1 has lunch 3, child 2-lunch 1, and child 3-lunch 2. The numbers of the children and lunches are different, so once again, no child selected his own.
Since child 1 has no more options, whoever he happens to be, we exhausted all possibilities. There are two ways therefore in which the lunches can be selected:

Choice 1
Child 1- Lunch 2; Child 2- Lunch 3; Child 3 - Lunch 1;

Choice 2

Child 1 - Lunch 3; Child 2- Lunch 1; Child 3- Lunch 2

So, we specified how the choices look, but the bottom line is the answer, and it says that there are exactly 2 ways, the nature of which is shown above.
Hope that helps

Answer 3

1x1x1 x3

child one's choices X child two's choices X child three's choices x the number of different children.

There are three ways

The first child has to have his own. The second child can only have child number three's lunch, since number one has his own. And that leaves only number two's lunch for three to take.

In the next case, there would still only be one outcome, but #2 would have his own lunch

And in the final case there is only one combination, making a third possible way, child 3 having thir own lunch.

Answer 4

suppose the children r called a-b-c, then 2nd way is b-c-a, 3rd way is c-a-b