SOLUTION: 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?
A) 6 (B) 12 (
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-> SOLUTION: 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?
A) 6 (B) 12 (
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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?
A) 6 (B) 12 (C) 14 (D) 16 (E) 24 Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! It would be 24 if there were no restrictions
For each in first position there are 6 ways 4*3*2*1
(abcd) bacd cabd (dabc)
abdc badc cadb (dacb)
(acbd) (bcad) (cbda) dbac
acdb (bcda) (cbad) (dbca)
(adbc) bdac cdab dcab
(adcb) bdca cdba (dcba)
With the restriction we have 2+4+4+2=12
Brad and Cathy can't be next to each other.
All combos of bc and cb are eliminated
Either in 1st and 2nd position
bcad bcda
cbad cbda
either in 2nd and 3rd
abcd dbca
acbd dcba
either in 3rd and 4th
adbc dabc
adcb dacb
That removes 12 leaving 12
