Solution:

To find the number of arrangements of the word SWORD such that no letter is in its original position, we can use the principle of inclusion-exclusion.

Let A1 be the event that the first letter is in its original position, A2 be the event that the second letter is in its original position, and so on. Then, we want to find the number of arrangements that satisfy none of these events.

Using inclusion-exclusion, the number of arrangements that satisfy at least one of these events is given by:

|A1 ∪ A2 ∪ A3 ∪ A4 ∪ A5| = |A1| + |A2| + |A3| + |A4| + |A5| - |A1 ∩ A2| - |A1 ∩ A3| - |A1 ∩ A4| - |A1 ∩ A5| - |A2 ∩ A3| - |A2 ∩ A4| - |A2 ∩ A5| - |A3 ∩ A4| - |A3 ∩ A5| - |A4 ∩ A5| + |A1 ∩ A2 ∩ A3| + |A1 ∩ A2 ∩ A4| + |A1 ∩ A2 ∩ A5| + |A1 ∩ A3 ∩ A4| + |A1 ∩ A3 ∩ A5| + |A1 ∩ A4 ∩ A5| + |A2 ∩ A3 ∩ A4| + |A2 ∩ A3 ∩ A5| + |A2 ∩ A4 ∩ A5| - |A1 ∩ A2 ∩ A3 ∩ A4| - |A1 ∩ A2 ∩ A3 ∩ A5| - |A1 ∩ A2 ∩ A4 ∩ A5| - |A1 ∩ A3 ∩ A4 ∩ A5| - |A2 ∩ A3 ∩ A4 ∩ A5| + |A1 ∩ A2 ∩ A3 ∩ A4 ∩ A5|

Now, we know that |A1| = 1 (since the first letter must be in its original position), and similarly for each of the other events. We also know that |A1 ∩ A2| = 1 (since both the first and second letters must be in their original positions), and similarly for each of the other intersections. The only unknown term is |A1 ∩ A2 ∩ A3 ∩ A4 ∩ A5|, which represents the number of arrangements where all five letters are in their original positions.

To find this term, we can use the principle of inclusion-exclusion again. Let B1 be the event that the first two letters are in their original positions, B2 be the event that the first three letters are in their original positions, and so on. Then, we have:

|A1 ∩ A2 ∩ A3 ∩ A4 ∩ A5| = |B1| - |B2| + |B3| - |B4| + |B5|

Again, we can easily compute each of the terms on the right-hand side. For example,

No of derrangements =n!(1/2! -1/3!+ 1/4! -..(-1)^n(1/n!)=5!(1/2!-1/3!+..-1/5!)=120(60-20+5-1)/120=44