there are digits from 1 to 6 .from that 5 digit no are to be formed such that they are divisible by 4. find no of possible combinations?  Answers
A) 1944 ways.  B) 1844 ways.  C) 1744 ways.  D) 1644 ways.  E) 1544 ways.
A) w/o repeatations.   possible last two digits= 12, 16, 24, 32, 36, 52, 56, 64   so total of 8 ways to fill last two digit.   remaining 3 digits can be filled by  (6-2)= 
<gadget id="calculator"> (6-2)</gadget>
<output>4</output>
4! ways=24 ways   so total no of ways=24*8= 
<gadget id="calculator">24*8</gadget>
<output>192</output>
192 ways.   B)with repeataions   possible last two digits= 12, 16, 24, 32, 36,44, 52, 56, 64   so total of 9 ways to fill last two digit.   remaining 3 digits can be filled by  6*6*6= 
<gadget id="calculator"> 6*6*6</gadget>
<output>216</output>
216 ways   so total no of ways=216*9= 
<gadget id="calculator">216*9</gadget>
<output>1_944</output>
1944 ways.   1944 ways.NSWER:1944 ways.
<result>A</result>
there are digits from 1 to 6 .from that 5 digit no are to be formed such that they are divisible by 4. find no of possible combinations?