JEE Advance - Mathematics (2020 - Paper 2 Offline - No. 13)
An engineer is required to visit a factory for exactly four days during the first 15 days of every month and it is mandatory that no two visits take place on consecutive days. Then the number of all possible ways in which such visits to the factory can be made by the engineer during 1-15 June 2021 is ...........
Answer
495
Explanation
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
$${}^{8 + 5 - 1}{C_{5 - 1}} = {}^{12}{C_4} = {{12 \times 11 \times 10 \times 9} \over {4 \times 3 \times 2}} = 495$$
$$ \therefore $$ 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
$${}^{8 + 5 - 1}{C_{5 - 1}} = {}^{12}{C_4} = {{12 \times 11 \times 10 \times 9} \over {4 \times 3 \times 2}} = 495$$
Comments (0)
