There are $13$ letters of $8$ different sorts $I,I,I,S,S,T,T,L,L,A,O,N,D$. In finding groups of $4$, how many permutations can be made if following are the possibilities to be considered? If $2$ are alike of one kind and $2$ are alike of other kind.

To form groups of 4 with 2 alike of one kind and 2 alike of another, we must select 2 letter types from the 4 available types that have repetitions (I, S, T, L). The number of ways to choose these 2 types is 4C2 = 6. For each chosen pair of letter types, there is exactly 1 way to select the letters, and they can be arranged in 4! / (2! * 2!) = 6 permutations. Multiplying these gives 6 * 6 = 36.

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