Question

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 1

we have to find the number of all possible ways in which such visits to the factory can be made by the engineer during 1 - 15 June 2021

solution : it can be solved easily. how ? let's solve it.

An engineer has to visit the factory for exactly four days during the first 15 days.

so, number of remaining days = 15 - 4 = 11 days

number of gape = number of remaining days + 1

= 11 + 1 = 12

now we have to arrange 4 days in 12 gape i.e., ¹²C₄ = 12!/(4! × 8!)

= 12 × 11 × 10 × 9/24

= 11 × 5 × 9

= 495

Therefore 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 495

Answer 2

Answer:

we have to find the number of all possible ways in which such visits to the factory can be made by the engineer during 1 - 15 June 2021

solution : it can be solved easily. how ? let's solve it.

An engineer has to visit the factory for exactly four days during the first 15 days.

so, number of remaining days = 15 - 4 = 11 days

number of gape = number of remaining days + 1

= 11 + 1 = 12

now we have to arrange 4 days in 12 gape i.e., ¹²C₄ = 12!/(4! × 8!)

= 12 × 11 × 10 × 9/24

= 11 × 5 × 9

= 495

Therefore 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 495

Step-by-step explanation:

$mark \;me\; as \;the \;brainliest \;please$