Correct Answer - Option 3 :
\(\dfrac{9!}{2!\times 2!}\)
Concept:
Suppose a set of n objects has n1 of one kind of object, n2 of a second kind, n3 of a third kind, and so on with n = n1 + n2 + n3 +...+ nk then the number of distinguishable permutations of the n objects is = \(\rm \frac{n!}{n_1!n_2! ...n_k!}\)
Calculation:
The given word is MATHEMATICS
The total number of letters is 11.
Given that we have to form those words which start and end with T.
T _ _ _ _ _ _ _ _ _ _ T
Since there are two identical letters and have to be placed in two positions so we can place it only 1 way.
Now we have the remaining 9 positions and the remaining 9 letters are M, A, H, E, M, A, I, C, S
Now in the given word, there are 2T’s, 2M’s and 2A’s, from where 2T’s are already used, so since there are 2M’s and 2A’s
Therefore, the total number of words = \(\dfrac{9!}{2!\times 2!}\)