QJEEAdv20Ma8983: JEE Advanced 2020 Mathematics - Class 11 JEEAdv20Ma8983 Solution: An engineer is required to visit a factory for exactly four…

Let the engineer visits the factory first time after x₁ days to 1 June, second time after x₂ days to first visit and so on.

$$\therefore$$ x₁ + x₂ + x₃ + x₄ + x₅ = 11

where x₁, x₅ $$\ge$$ 0 and x₂, x₃, x₄ $$\ge$$ 1 according to the requirement of the question.

Now, let x₂ = a + 1, x₃ = b + 1 and x₄ = c + 1 where a, b, c $$\ge$$ 0

$$\therefore$$ New equation will be

x₁ + a + b + c + x₅ = 8

Now, the number of all possible ways in which the engineer can made visits is equals to the non-negative integral solution of equation

x₁ + a + b + c + x₅ = 8, and it is equal to

$${}^{8+5-1}{C}_{5-1}={}^{12}{C}_4=\frac{12\times 11\times 10\times 9}{4\times 3\times 2}=495$$