There are 11 letters, of which I appears 4 times, S appears 4 times, P appears 2 times & M appears 1 time. ∴ The required number of arrangements = $\frac{11!}{4!4!2!}$ = $\frac{11\times10\times9\times8\times7\times6\times5\times4!}{4\times3\times2\times2\times4!}$ = 11 × 10 × 9 × 7 × 5 = 34650 ... (i) When four I’s come together, we treat them as a single object. This single object with 7 remaining objects will account for 8 objects. These 8 objects in which there are 4S’s & 2P’s can be rearranged in $\frac{8!}{4!2!}$ ways i.e., in 840 ways ... (ii) Number of arrangements when four I’s do not come together = 34650 – 840 = 33810.