document.write( "Question 775611: 44. A, B, C, D, E and F are invited to attend a meeting. These 6 wacky guys put forward the following requirement:
\n" ); document.write( "1) At least 1 person out of A and B is going to attend the meeting
\n" ); document.write( "2) 2 persons out of A, E and F are going to attend the meeting
\n" ); document.write( "3) B and C decided to either attend the meeting together or abandon the chance together
\n" ); document.write( "4) 1 person out of A and D is going to attend the meeting
\n" ); document.write( "5) 1 person out of C and D is going to attend the meeting
\n" ); document.write( "6) If D decides not to attend the meeting, E will do the same
\n" ); document.write( "Could you tell us who of them finally attend the meeting?
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Algebra.Com's Answer #473026 by psbhowmick(878)\"\" \"About 
You can put this solution on YOUR website!
Case I: A attends the meeting
\n" ); document.write( "D will not attend the meeting (rule 4)
\n" ); document.write( "C will attend the meeting (rule 5)
\n" ); document.write( "E will not attend the meeting (rule 6)
\n" ); document.write( "F will attend the meeting (rule 2)
\n" ); document.write( "B will attend the meeting (rule 3)\r
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\n" ); document.write( "\n" ); document.write( "Case II: A does not attend the meeting
\n" ); document.write( "E and F will attend the meeting (rule 2)
\n" ); document.write( "D will attend the meeting (rule 4)
\n" ); document.write( "C will not attend the meeting (rule 5)
\n" ); document.write( "B will not attend the meeting (rule 3)
\n" ); document.write( "But if A and B both do not attend the meeting then rule 1 is violated.
\n" ); document.write( "Therefore, the case A does not attend the meeting cannot be true.
\n" ); document.write( "Hence, Case I is feasible.\r
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\n" ); document.write( "\n" ); document.write( "So, A, B, C and F will finally attend the meeting.
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