document.write( "Question 293160: Four children Alice, Brad, Cathy and Dan are arranged in a line. If Brad and Cathy cannot be next to each other, in how many ways can the kids be arranged?\r
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document.write( "A) 6 (B) 12 (C) 14 (D) 16 (E) 24 \n" );
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Algebra.Com's Answer #211613 by richwmiller(17219)![]() ![]() You can put this solution on YOUR website! It would be 24 if there were no restrictions \n" ); document.write( "For each in first position there are 6 ways 4*3*2*1 \n" ); document.write( "(abcd) bacd cabd (dabc) \n" ); document.write( "abdc badc cadb (dacb) \n" ); document.write( "(acbd) (bcad) (cbda) dbac \n" ); document.write( "acdb (bcda) (cbad) (dbca) \n" ); document.write( "(adbc) bdac cdab dcab \n" ); document.write( "(adcb) bdca cdba (dcba) \n" ); document.write( "With the restriction we have 2+4+4+2=12 \n" ); document.write( "Brad and Cathy can't be next to each other. \n" ); document.write( "All combos of bc and cb are eliminated \n" ); document.write( "Either in 1st and 2nd position \n" ); document.write( "bcad bcda \n" ); document.write( "cbad cbda \n" ); document.write( "either in 2nd and 3rd \n" ); document.write( "abcd dbca \n" ); document.write( "acbd dcba \n" ); document.write( "either in 3rd and 4th \n" ); document.write( "adbc dabc \n" ); document.write( "adcb dacb \n" ); document.write( "That removes 12 leaving 12 \n" ); document.write( " \n" ); document.write( " |