how many different ways can 2 students be seated in a row of 4 desks, so that there is always at least one empty desk between the students?  Pick:
A) 2
B) 3
C) 4
D) 6
E) 12
<gadget id="calculator">4 - 1</gadget>
<output>3</output>

<gadget id="calculator">factorial(3) / factorial(3 - 2)</gadget>
<output>6</output>

<result>D</result>
how many different ways can 2 students be seated in a row of 4 desks, so that there is always at least one empty desk between the students?
permutation(subtract(4, const_1), 2)
"total cases : 12 ( student one has 4 options and student two has three options , 4 x 3 = 12 ) non - favourable cases : 6 ( when two students sit together . students in desk 1 and desk 2 , in desk 2 and desk 3 , in desk 3 and desk 4 ) for each of these cases there are two possibilities because the positions can be interchanged . hence 2 x 3 = 6 . so favourable cases : 12 - 6 = 6 . answer : d"