find the number of permutations of letters of CALENDAR in which C and A are together but N and D are not?
That's
the number of ways C&A are together
minus
the number or ways C&A are together and N&D are together
So:
N(C&A together) = number of permutations of these 7 things:
{CA,L,E,N,D,A,R), which is 7!
plus the number of permutations of these 7 things:
{AC,L,E,N,D,A,R), which is also 7!
So that's 2*7!
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N(C&A together and N&D together) =
the number of permutations of these 6 things:
{CA,L,E,ND,A,R} which is 6!
plus
the number of permutations of these 6 things:
{CA,L,E,DN,A,R} which is 6!
plus
the number of permutations of these 6 things:
{AC,L,E,ND,A,R} which is 6!
plus
the number of permutations of these 6 things:
{AC,L,E,DN,A,R} which is 6!
which is 4*6!
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Therefore the desired number is
2*7! - 4*6! = 2*5040 - 4*720 = 10080 - 2880 = 7200
Edwin
