How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D? Choose the correct choice.
A) 96	B) 60	C) 48	D) 36	E) 17
Total # of permutation of 5 distinct letters will be 5!=120;   Glue A and B together, consider it to be one unit: {AB}{C}{D}{E} --&gt; # of permutation of these 4 units will be 4!=24, A and B within its unit also can be arranged in 2 ways : {AB} or {BA}, so total # of ways to arrange A, B, C, D, and E so that A and B to be together will be 4!*2=48;   The same for A and D: total # of ways to arrange A, B, C, D, and E so that A and D to be together will be 4!*2=48;   Now, the above   48+48= 
<gadget id="calculator">  48+48</gadget>
<output>96</output>
96 cases will contain the arrangements when A is adjacent to both B and D, so we should subtract this cases to get rid of the double counting. The # of case when A is adjacent to both B and D will be: consider {BAD} {C}{E} --&gt; # of permutation of these 3 units will be 3!, {BAD} also can be arranged in 2 ways: {BAD} or {DAB}, so total # of ways to arrange A, B, C, D, and E so that A is adjacent to both B and D will be 3!*2=12;   The # of arrangements when A is adjacent to neither B nor D will be total-(48+48-12)=120-84= 
<gadget id="calculator">120-84</gadget>
<output>36</output>
36.   Answer: 36.
<result>D</result>
How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?