problem 1

There are

(\frac{11}{4})

ways to select $4$ positions for $I$ ,

(\frac{7}{4})

ways to select $4$ postions for $S$ ,

(\frac{3}{2})

ways to selection $2$ positions for $P$ leaving us with a single choice of position for $M$ . In total, we get

(\frac{11}{4}) (\frac{7}{4}) (\frac{3}{2}) (\frac{1}{1})

permutations.