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:
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document.write( "1) At least 1 person out of A and B is going to attend the meeting
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document.write( "2) 2 persons out of A, E and F are going to attend the meeting
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document.write( "3) B and C decided to either attend the meeting together or abandon the chance together
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document.write( "4) 1 person out of A and D is going to attend the meeting
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document.write( "5) 1 person out of C and D is going to attend the meeting
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document.write( "6) If D decides not to attend the meeting, E will do the same
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document.write( "Could you tell us who of them finally attend the meeting?
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document.write( "
Algebra.Com's Answer #473026 by psbhowmick(878) 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 \n" ); document.write( " \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 \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So, A, B, C and F will finally attend the meeting. \n" ); document.write( " |