How many ways we can arrange A,B,C,D letters so that AB,BC,CD should not be together

The letters A,B,C,D can be arranged in $4!$ ways.
Total possible ways of arranging the letters are $4!=24$
In order to calculate neither a and b nor c and d come together,first we have to see the cases when they do come togther
So we take a and b as a single element,due to which the total elements are now 3
$\therefore$ 3 letters can be arranged in $3!$ ways and a and b itself can be arranged in $2!$.
Cases when a and be are together can be arranged in $3!*2!$ ways
$i.e,3!*2!\rightarrow3*2*1*2*1=12\:$ ways
Similarly for the case when c and d are together i.e, 12 ways.
Now the case when a and b and c and d are together,the case becomes
$2!*2!*2!=2*2*2=8$ ways
So the total number of ways when neither of them are together are
$=24-12-12+8=8$
Hence there are 8 number of ways when letters a b c and d can be arrange in which neither of them are together.

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