There are eight boxes of chocolates, each box containing distinct number of chocolates from 1 to 8. In how many ways four of these boxes can be given to four persons (one boxes to each) such that the first person gets more chocolates than each of the three, the second person gets more chocolates than the third as well as the fourth persons and the third person gets more chocolates than fourth person?
A) 70 B) 40 C) 72 D) 80
Answers
Answered by
2
72.....................................
Answered by
1
Answer:
Step-by-step explanation:
All the boxes contain distinct number of chocolates.
For each combination of 4 out of 8 boxes, the box with the greatest number has to be given to the first person, the box with the second highest to the second person and so on.
The number of ways of giving 4 boxes to the 4 person is,
8C4 = 70
Similar questions