How to solve Probability How many ways can four people be seated in a row of four seats? I tried 4 times 4 which equals 16 but im not so sure about this one i dont know how to solve it?
 A: Is this homework for a class you're currently studying? These kinds of questions usually get taught in this manner:
There are 4 people who can go in the first seat.
Having chosen 1 of those people, there are 3 people left who can go in the second seat.
Having chosen 1 of them, there are now 2 people left for the third seat.
There is then only 1 person left for the final seat.
Because you need to combine all of these possibilities, there are 4x3x2x1 = 24 different ways of arranging the people into the seats. You can also show that this is the case by just listing all the possibilities - let's say our four people are named A, B, C and D. Then they can sit like this:
ABCD ABDC ACBD ACDB ADBC ADCB
BACD BADC BCAD BCDA BDAC BDCA
CABD CADB CBAD CBDA CDAB CDBA
DABC DACB DBAC DBCA DCAB DCBA
And by looking at those you can see the way our initial thinking worked - each row shows how to fix one of the 4 people in the first seat, then arrange the remaining 3 in the other seats.
A: Say their names are $A$, $B$, $C$, and $D$.
Any of the four can sit in the first seat.
If $A$ is in the first seat, any of three could sit in the second seat, so we could have $AB$, $AC$, or $AD$.
If $B$ is in the first seat, any of three could sit in the second seat, so we could have $BA$, $BC$, or $BD$.
If $C$ is in the first seat, any of three could sit in the second seat, so we could have $CA$, $CB$, or $CD$.
If $D$ is in the first seat, any of three could sit in the second seat, so we could have $DA$, $DB$, or $DC$.
So multiply $4$ by $3$.
Then consider the third seat.  If $AB$ are in the first two seats, we could have $ABC$ or $ABD$.  And so on: For each pair in the first two seats, we could have either of two in the third seat.
So think about $4\times3\times2$.
Then think about the fourth seat.
A: Well, let's say the four people are called Ann, Bari, Carl, Dave.
And the seats are 1, 2, 3, 4. 
If Ann sits in seat 1, then Bari, Carl, and Dave CAN'T sit in seat 1. 
So then, Bari sits in seat 2, then Carl and Dave CAN'T sit in seat 2. 
etc. 
The idea is that, if someone takes a seat, that seat is not available to anyone else. So let's start again. 
Ann, Bari, Carl, Dave have 4 options. Any 4 of them can sit in any seat right now. 
So let's say Ann gets to have first pick. Ann chooses some seat, and she sits down. There are only 3 options left for Bari, Carl, and Dave.
Let's say Carl chooses a seat, now there are only 2 options left for Bari and Dave.
Bari chooses a seat, so now Dave has to choose the last remaining seat. 
We multiply the options together because it does not matter who sits where, just that they are all seated in the end
It starts with 4 options in the first seat. Then there were 3 options in the second, then there were 2 in the third, then there was 1 in the last. So 4 * 3 * 2 * 1 = 24 options. 
A: Let us start with two people, A and B.
There are $2$ orders that they can be seated in a row of $2$ seats: AB or BA.
Now consider $3$ people, A, B, and C, and a row of $3$ seats. They can be seated in the orders ABC, ACB, BAC, BCA, CAB, CBA. This gives a total of $6$ ways.
Now that we are warmed up, let's deal with $4$ people, A, B, C, and D, and $4$ seats.
We start listing. In order not to make any mistakes, we list systematically. A seating corresponds to a $4$-letter "word" in the alphabet A, B, C, D, where each word uses each letter exactly once.  Let us list these words in alphabetical order.  
There are the words that start with A: they are ABCD, ABDC, ACBD, ACDB, ADBC, ADCB. That gives $6$.
There are the words that start with B. There is no real need to list them, by symmetry there are just as many symmetry there are just as many there are  words  begin with A, $6$ of them.
And there are exactly $6$ that begin with C, and $6$ that begin with D, a total of $24$.
So far we have $2$ ways for $2$ people, $6$ ways for $3$ people, and $24$ ways for $5$ people.
Let us deal with $5$ people, A, B, C, D, E, and $5$ seats. We count the $5$-letter "words."
Imagine listing them. There will be the words that begin with A, the ones that begin with B, and so on. How many begin with A? We need to append to A a $4$-letter word in the alphabet B, C, D, E. We know there are $24$ $4$-letter words, so there are $24$ $5$-letter words that begin with A. There are also $24$ $5$-letter words that begin with B, and $24$ that begin with each of C, D, E, for a total of $(5)(24)=120$.
Now many $6$ letter words in the alphabet A to F? Well, there are the ones that begin with A, the ones that begin with B, and so on. The ones that begin with A are obtained by a appending a $5$-letter word to A, so there are $120$. Same with the other initial letters, for a total of $(6)(120)$. And so on.

Let us start again. There are $2$ ways to seat $2$ people in a row.
Take $3$ people, A, B, C. There are $3$ ways to choose who goes into the leftmost chair. For each choice, the other $2$ people can be seated in $2$ ways, for a total of $(3)(2)$.
Take $4$ people. There are $4$ ways to choose who goes into the leftmost chair. For each choice, there are $(3)(2)$ ways to seat the other $3$ people in the remaining chairs, for a total of $(4)(3)(2)$.
Take $5$ people. The same reasoning gives $(5)(4)(3)(2)$ ways of seating them.
And so on.
A: 4 people for 4 seats. Seat 1 will have 4 choices of people. Seat 2 will have 3 choices of people after your have chosen for seat 1. Seat 3 will have 2 choices. Seat 4 has 1. It will be 4x3x2x1=4!
