Total letters = 12 in which N = 3, D = 2. E = 4.
Total number of Arrangements = n(S) =
12
!
3
!
.2
!
.4
!
12
(a) A : 4 E’s are together can be taken as I unit and remaining 8 totally 9 can be permuted in
9
!
3
!
.2
!
9
(b) B: 2D’s are together and 3N’s are together 1 + 1 + remaining 7 = 9 can be done in
(c) All 4E’s are not together = total – together = 1 – 0.0 18 = 0.982
(d) D: No. two Es are together
:. E’s are in between the other B letters, there are 9 spaces, in which 4 Es are placed in 9C4 ways and 8 letters can be permuted in
8
!
3
!
.2
!
8
and 4E’s are repeated need to delete n(D) = 9C4
8
!
3
!
.2
!
.4
!
8
