Raginsheep
Active Member
- Joined
- Jun 14, 2004
- Messages
- 1,227
- Gender
- Male
- HSC
- 2005
Assuming you have the letters AA BB CC DD, in how many ways can you arrange them so no two letters of the same type are together?
-
Looking for HSC notes and resources? Check out our Notes & Resources page
Perms and Combs Question (1 Viewer)
- Thread starter Raginsheep
- Start date
Raginsheep
Active Member
- Joined
- Jun 14, 2004
- Messages
- 1,227
- Gender
- Male
- HSC
- 2005
The answer is 864. Apparently.
grendel
day dreamer
- Joined
- Sep 10, 2002
- Messages
- 103
- Gender
- Male
- HSC
- N/A
ok her goes.
the total number of arrangements=(total with no restrictions)-
..........................[(total arrangements with all four letters adjacent)+
.............................(total arrangements with three of the letters adjacent)+
...............................(total arrangements with two of the letters adjacent)+
.................................(total arrangements with one letter adjacent)]
total with no restrictions = 8!/(2!2!2!2!)=2520
total arrangements with all four letters adjacent=4!=24
total arrangements with three of the letters adjacent=[5!/2!-24] X 4C3=144
total arrangements with two of the letters adjacent=[6!/(2!2!)-(24+36+36)] X 4C2=504
total arrangements with one letter adjacent=[7!/(2!2!2!)-(24+36+36+36+84+84+84)] X 4C1=984
TF the total number of arrangements = 2520-(24+144+504+984) = 864
NOTE: all numbers in red are subtracted from the total as these were counted in the previous case(s)
all numbers in the [] are arrangements for one instance only and are multiplied by the numbers in blue to take into account all the different instances.
the total number of arrangements=(total with no restrictions)-
..........................[(total arrangements with all four letters adjacent)+
.............................(total arrangements with three of the letters adjacent)+
...............................(total arrangements with two of the letters adjacent)+
.................................(total arrangements with one letter adjacent)]
total with no restrictions = 8!/(2!2!2!2!)=2520
total arrangements with all four letters adjacent=4!=24
total arrangements with three of the letters adjacent=[5!/2!-24] X 4C3=144
total arrangements with two of the letters adjacent=[6!/(2!2!)-(24+36+36)] X 4C2=504
total arrangements with one letter adjacent=[7!/(2!2!2!)-(24+36+36+36+84+84+84)] X 4C1=984
TF the total number of arrangements = 2520-(24+144+504+984) = 864
NOTE: all numbers in red are subtracted from the total as these were counted in the previous case(s)
all numbers in the [] are arrangements for one instance only and are multiplied by the numbers in blue to take into account all the different instances.
justchillin
Member
- Joined
- Jun 11, 2005
- Messages
- 210
- Gender
- Male
- HSC
- 2005
Lesson learnt: i hate probability and counting... 