Eight boxes contain distinct numbers of chocolates from 1 to 8. In how many ways can four boxes be given to four persons, one box to each, so that the first person receives more chocolates than each of the other three, the second person receives more than the third and fourth persons, and the third person receives more than the fourth person?

Introduction / Context:
This problem combines basic combinatorics with ordering constraints. You have boxes with distinct quantities of chocolates and you want to distribute four of them to four labelled persons in such a way that their chocolate counts follow a strict decreasing order from person one to person four. Understanding how to convert such inequality conditions into a simple counting argument is a useful exam skill.

Given Data / Assumptions:

• There are 8 boxes, containing 1, 2, 3, 4, 5, 6, 7 and 8 chocolates respectively.
• Four persons receive one box each.
• Let their chocolates be x1, x2, x3 and x4 for persons 1, 2, 3 and 4 respectively.
• The conditions are: x1 greater than x2, x3 and x4; x2 greater than x3 and x4; and x3 greater than x4.

Concept / Approach:
The inequalities force a strict ordering: x1 is the largest, x2 is the second largest, x3 is the third largest and x4 is the smallest of the four chosen numbers. Once you choose any set of four distinct box sizes, the assignment to persons is completely fixed by these conditions. Therefore, the count reduces to choosing 4 box sizes out of 8, without regard to order: this is simply 8C4.

Step-by-Step Solution:
We need to select any 4 distinct numbers out of 8: compute 8C4.Use formula: 8C4 = 8 * 7 * 6 * 5 / (4 * 3 * 2 * 1).Simplify: numerator 8 * 7 * 6 * 5 = 1680; denominator 24.Compute 1680 / 24 = 70.For each such set, the largest goes to person 1, second largest to person 2, third to person 3, and smallest to person 4.

Verification / Alternative check:
Instead of thinking about assignments, note that any 4 chosen sizes can be uniquely arranged to satisfy the required inequalities.There is exactly 1 valid assignment for each 4 element subset of the 8 box sizes.Thus the final count equals the number of such subsets, which is 8C4 = 70.

Why Other Options Are Wrong:
40, 72 and 80 do not correspond to any meaningful combination count for selecting 4 of 8 under these constraints.Some of these values may come from misapplying permutations or forgetting that the inequalities already fix the order.

Common Pitfalls:
Trying to permute the four chosen numbers among the four people even though the inequalities already fix who gets which size.Misinterpreting the condition as only requiring x1 to be maximum and ignoring the full strict chain x1 greater x2 greater x3 greater x4.Computing permutations 8P4, which is much larger and incorrect.

Final Answer:
The number of valid distributions is 70.