How many ways to arrange ABCD such that A precedes B, and C precedes D

Question

Restriction 1: A precedes B

Restriction 2: C precedes D

Note: They can be either adjacent or not.

For example,

ABCD

ACBD

ACDB

CABD

CDAB

but

CBAD

ABDC

are not allowed and so on, as long as A comes before B and C comes before D.

Is there a generalized way to solve such problems?

Another similar problem to the above is finding the number of arrangements of ABCDE if

Restriction 1: A precedes B, B precedes C

Restriction 2: D precedes E

e.g.

ABCDE

ADBCE

DEABC

Answer 1

First we notice there are not restrictions on $B,D$ so we place them down. $$B,D$$ There are 2 ways to rearrange this.

Now we'll consider our first restriction for $A$ the only possible place we can add A is before B so now our list looks like this. $$A,B,D$$ There are still only 2 ways since $A$ has only one possible position.

Now we'll our second restriction. Notice how $C$ can be placed in 3 possible places: Between $BD,AB$ and before $A$. so we end up with $$A,B,C,D$$ which has $(2!)(1!)(3!)$ ways of rearranging.

In general you can go through this process for any permutation with restrictions as long as you start with the elements that have no restrictions on them.

Answer 2

There are in total $4!=24$ ways to arrange $ABCD$. Exactly half of them will have $B$ before $A$, and exactly half of them will have $D$ before $C$. Since the relative ordering of $A$ and $B$ is clearly independent of the relative ordering of $C$ and $D$, under those restrictions, the total number of ways to arrange the items is $24/(2\cdot 2) = 6$.

Answer 3

I try to solve it by generalize method so that we can apply this for other cases also.

We A, B, C, D.

If we fix A on first place and B on last place. Then second and third place can be filled in 1! * 1!. So number of ways = 1 * 1! * 1! * 1 = 1

So if the pair is on first and last place so we have only 1 way. So for C is on first place and D on last place we have 1 way.

If pair is not on first and last place. Then,

A is on first place and D. Either B is on second or third. And third position is filled with remaining.

Ways = 1 * 2! * 1! * 1 = 2

Similarly for C is on first and B on last. Then,

Ways = 1 * 2! * 1! * 1 = 2

Total ways = 1 + 1 + 2 + 2 = 6.

Note -

So if n pairs then if pair elements on first and last position then ways n * (n-1)!

If pair element not on first and last place then ways n * n!

So total ways = n * (n-1)! + n * n!