In an examination, 5 students have been allotted their seats as per their roll numbers. The number of ways, in which none of the students sits on the allotted seat, is _________.

Solution

Maths sir · 4 steps

Nobody in their own seat is a derangement: use inclusion-exclusion on the 5! seatings.

1. Name the count. Arranging 5 students so that no one gets their own seat is a derangement of 5 objects, written D₅.

   Dₙ = n!(1 - 1/1! + 1/2! - 1/3! + … + (-1)ⁿ/n!)

2. Where it comes from. Start with all 5! = 120 seatings. Subtract those with at least one student in the right seat, add back those with at least two, and keep alternating. That is inclusion-exclusion, and it gives the formula above.
3. Put in numbers. D₅ = 120(1/2 - 1/6 + 1/24 - 1/120) = 60 - 20 + 5 - 1 = 44. The recurrence Dₙ = (n-1)(Dₙ₋₁ + Dₙ₋₂) agrees: D₅ = 4(9 + 2) = 44.
4. So. The number of ways is 44.

The common slip: 119. Subtracting only the one seating where everyone is correct gives 120 - 1 = 119. That counts "not all correct", not "none correct".

The idea to keep. When nothing may stay in its own place, think derangement. Learn D₁ to D₅ by heart: 0, 1, 2, 9, 44.

Answer from the NTA final answer key. Solution by PrepWiser.