In an examination 10,000 students appeared. The result revealed the number of…

Concept — Counts of the form “exactly k of n conditions satisfied” form a partition: every individual falls into exactly one such group, so groups for different values of k never overlap and can simply be added. Two consequences drive this item. First, “at least k” is the plain sum of the disjoint groups “exactly k”, “exactly k + 1”, …, “exactly n”. Second, failing exactly one of n conditions is the very same event as satisfying exactly n − 1 of them.

Application — Here n = 5 subjects. Each “failed in one named subject only” line counts the students who lost that single subject and therefore cleared the other four. A student can appear on only one such line, so these five tallies are disjoint and add up to the group that passed exactly four subjects.

1. Students who passed exactly four subjects = 75 + 145 + 140 + 200 + 157 = 717.

2. Students who passed all five subjects = 5,583, read directly from the data.

3. “At least four subjects” covers precisely these two disjoint groups — exactly four, and exactly five — so the required count is their sum.

4. Required count = 717 + 5,583 = 6,300.

Cross-check — Add every “exactly k” group: 5,583 (five) + 717 (four) + 1,400 (three) + 1,200 (two) + 735 (one) = 9,635. Out of the 10,000 students who appeared, that leaves 10,000 − 9,635 = 365 students who passed no subject at all. The remainder is non-negative and the categories account for the whole cohort exactly once, so nothing has been double-counted or dropped, and the total of 6,300 stands.

Result — 6,300 students passed in at least four subjects.