How many ways can four letters abcd be arranged such that a always comes before b and c always comes before d?
Total number of ways abcd can be arranged? 4!
Half of them a if before b, half b is before a.
Similarly, in half, c is before d, and other half, d is before a.
But we can't divide 4! by 2 – because there are some in which b is before a AND d is before c. So what's the answer?
Best Answer
It would be $\frac{4!}{2\cdot 2}$. The "a before b" and "c before d" are two independent events, so we can straightaway divide by two twice. You can also look at it this way:
There are four (equally likely) possibilities that can occur when we arrange a,b,c,d without any rules --
Only one out of the four equally likely possibilities is favourable, and we have a total number of $4!$ cases, thus the answer if $\frac{4!}{4}=6$. If you want, the favourable cases are: abcd,acbd,acdb,cdab,cabd,cadb.