(c) \(rac{1}{8}\)Let S be the sample space.Then n(S) = Total number of ways in which the letters of word NATIONAL can be arranged= \(rac{8!}{2!\,2!}\) (∵There are 2A‘s and 2N's in 8 letters)Let E : Event of arranged letters of the word NATIONAL, so that the last position is occupied by letter ‘T‘. Here position of T is fixed, so n(E) = Number of ways in which the rest of the 7 letters can be arranged in 7 places.= \(rac{7!}{2!\,2!}\) (∵ There are 2A‘s and 2N's)∴ Required probability = \(rac{n(E)}{n(S)}\) = \(rac{rac{7!}{2!\,2!}}{rac{8!}{2!\,2!}}\) = \(rac{7!}{8!}\) = \(rac{7!}{8 imes7!}\) = \(rac{1}{8}\).