How many different combinations can you get with 5 people standing in line?

Hey!

This problem can be solved using permutations, but to make it easier to understand, I will explain it in simple terms.

Basically, there are five spots in the line. Let us look at each spot individually.

[Spot 1][Spot 2][Spot 3][Spot 4][Spot 5]

Now, in spot 1, how many people can possibly be placed there? This answer would be 5 because there is a total of five people. Let us note this number in our small diagram:

[5 choices][Spot 2][Spot 3][Spot 4][Spot 5]

For the second spot, the same question should be asked. How many people can possibly be placed here? This answer would be 4, because one person has already been placed in spot one. Thus, we fill in this number:

[5 choices][4 choices][Spot 3][Spot 4][Spot 5]

This next statement informs you of what to do with this numbers that are being obtained. Note that for each of the five choices in spot one, there are four choices that can be placed in spot two.

"For each" is the key term and means that we must multiply.

If we fill out the rest of the chart, the final product would be:

[5 choices][4 choices][3 choices][2 choices][1 choice]

For each of the four choices in the second spot, three choices are available for the third spot.

For each of the three choices in the third spot, two choices are available for the fourth spot.

For each of the two choices in the fourth spot, one choice is available for the fifth spot.

Thus, we can make an expression, using multiplication:

5 * 4 * 3 * 2 * 1

The above expression is known as a factorial. A factorial is the multiplication of all integers less than the factorial number, but greater than or equal to 1. Note that the factorial number must be a positive integer.

The above expression is 5! (! is the sign for factorial). If you multiply the expression out, the answer is obtained.

5!

= 5 * 4 * 3 * 2 * 1

= 20 * 3 * 2 * 1

= 60 * 2 * 1

= 120 * 1

= 120

Thus, 120 is the answer. This is how many different possible combinations there are.

I hope this helps!

This is a commonly used to determine the number of possible combinations given b number of different types.

Lets take a simple example which is similar. Suppose we are using binary, so each number can have 2 possible combinations 0 or 1.

Lets figure out how many binary combinations we can have for 2, 3, or 4

This works by making the base the number of possible combinations and the exponent the number of places.

so our answers are

2^2, 2 ^3, 2 ^ 4

People are also binary if you consider that they are either male or female.

Here there are 5 people in a line, and 2 possibilities: male or female. Thus we have 2^5 = 64.

This way cares about the order if we only have 2 types of people.

An alternative which I think you mean is to suppose all 5 people are different.

Then each time we place one person in the line we have 1 less than before.

So we have

5 * 4 * 3 * 2 * 1 or 5! = 120

ok, we have five people ( a b c d e )

and five places in line ( 1 2 3 4 5 )

so let's count how many different combinations we can get:

at the place number 1 we can put any of our five people:

so we have 5 combo, ok.

now we have someone at no.1 and 4 persons left:

for position no.2 we can put any of these 4 left: so we get 4 combo. Together it is 5 * 4 = 20 combo for first two positions.

then for position no.3 we have 3 options: so in total 5 * 4 * 3

then for position no.4 we have 2 options left: so in total 5 * 4 * 3 *2 and finally only one person left for the last position.

so total number of combos is 5 * 4 * 3 * 2 * 1 = 120

and the general formula for this kind of problems is n!

(n factrorial)

n! = n* (n-1) *(n-2) * (n-3) ... * 2 * 1

Cheers!

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For each place in the line you multiply how many people can fill that spot. So for the first spot there are 5 choices, but for the next spot one of the people is already in the first spot, so there are 4 choices for the second spot and so on. This will be 5!.

you simply multiply five times itself because there are five people