Get Started With Combinatorics 1 – Alice, Bob, and Charlie

Let's list out the number of ways:

1. Alice, Bob, Charlie
2. Alice, Charlie, Bob
3. Bob, Alice, Charlie
4. Bob, Charlie, Alice
5. Charlie, Alice, Bob
6. Charlie, Bob, Alice

Hence, there are 6 orderings for three people to stand in line.

We are dealing with a 'permutation without repetition'

A 'permutation' is a combination wherein the ordering of objects matters.

For example, the ordering ' $Charlie, Alice, Bob$ ' is different from ' $Alice, Charlie, Bob$ '

We say 'without repetition' since no one can be both first and second in line at the same time.

For example, the ordering ' $Charlie, Charlie,Charlie$ ' is invalid since this would mean that Charlie has 2 clones standing in line with him.

The formula for permutations without repetitions is $\dfrac{n!}{(n-r)!}$ wherein $n$ is the number of objects

$r$ is the number of objects that you need and $n!=n*(n-1)*(n-2)*(n-3)*...*3*2*1$ For example, $5!=5*4*3*2*1$ (Note: $0!=1$ )

In this case $n=r=3$ since we have three objects: Alice, Charlie and Bob, and we are going to need all of the three of them because we are going to put all of them in a line

Substituting this in the formula mentioned earlier, we have $\frac{3!}{(3-3)!}\rightarrow\frac{3*2*1}{0!}\rightarrow\frac{6}{1}\rightarrow\boxed{6}$

if we cross product any set of 3 numbers with itself the no. of elements new set contain 3x3= 9 elements out of which 3ordered pair contain same elements so they are eliminated and remaning one element was added in each ordered pair and thus we get 6 orders for eg -A{123} so AXA={(123)(123)}={(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(32),(3,3)} ordered pair containing smae element=(1,1),(2,2),(3,3) these are eliminated remaining are=(1,2),(1,3),(2,1),(2,3),(3,1),(3,2) remaining elements in each ordered pair makes different orders=(1,2,3) (1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1) so there are six possible orders and thus answer is =6

For the person in front you can choose any one of them (3). For the person behind them you'll only have two choices, and that'll leave you with no choice for the one at the back. So it's 3 times 2 (times 1), or 3!.

The different orders are six in no. i. e: 1 Alice, Bob, Charlie.

2 Charlie, Bob, Alice.

3 Bob, Charlie, Alice.

4 Charlie, Alice, Bob.

5 Bob, Alice, Charlie.

6 Alice, Charlie, Bob.

What I do for these problems is square the number of choices(3 in this case), and subtract the number of choices from the result to remove the identicle pairs, if that makes sense. 3*3=9 9-3=6

3! Factorial way is proper to get the exact solution

3 people in row. so,the of ways, 3!=3 2 1=6

Answer :- 6

It's a permutation without repetitions. So The answer is 3!=6

Let's list the 3 people in line:

Alice A, Bob B, Charlie C

1. ABC
2. CAB
3. BCA
4. ACB
5. BAC
6. BCA

And that's 6 possible orders for those 3 to stand in line

2 people in line 2x1...3 in line 3x2x1

There it are 3 posibilities, then it are 2 posibilities and finally it are1 posibility. Hence 3+2+1=6