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
\n" ); document.write( "\n" ); document.write( "A) 6 (B) 12 (C) 14 (D) 16 (E) 24
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Algebra.Com's Answer #211613 by richwmiller(17219)\"\" \"About 
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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
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