Permutations & Combinations
Distribution
Grade 11
Question:
<p>In a three-storey building, there are four rooms on the ground floor, two on the first and two on the second floor. If the rooms are to be allotted to six persons, one person occupying one room only, the number of ways in which this can be done so that no floor remains empty is</p>
<p>(1) Options not fully visible</p>
<p>(2) Options not fully visible</p>
<p>(3) Options not fully visible</p>
<p>(4) Options not fully visible</p>
Step-by-Step Solution
Key Concept: Use inclusion-exclusion principle: Total ways minus ways where at least one floor is empty. First distribute 6 persons to 8 rooms, then subtract cases where ground floor OR first floor OR second floor has no one.
<p><strong>Step 1:</strong> Total ways to allot 6 persons to 8 distinguishable rooms (one person per room) = P(8,6) = 8!/(8-6)! = 8!/2! = 20,160</p><p><strong>Step 2:</strong> Apply inclusion-exclusion. Let A = ground floor empty, B = first floor empty, C = second floor empty.</p><p><strong>Step 3:</strong> |A| = persons allotted to 4 rooms (first + second) = P(4,6) = 0 (impossible, only 4 rooms available)<br>|B| = persons allotted to 6 rooms (ground + second) = P(6,6) = 6! = 720<br>|C| = persons allotted to 6 rooms (ground + first) = P(6,6) = 6! = 720</p><p><strong>Step 4:</strong> |A∩B| = persons allotted to 2 rooms (second only) = P(2,6) = 0 (impossible)<br>|A∩C| = persons allotted to 2 rooms (first only) = P(2,6) = 0 (impossible)<br>|B∩C| = persons allotted to 4 rooms (ground only) = P(4,6) = 0 (impossible)<br>|A∩B∩C| = 0</p><p><strong>Step 5:</strong> By inclusion-exclusion: Ways with no floor empty = 20,160 - (0 + 720 + 720) + 0 - 0 = 20,160 - 1,440 = 18,720</p><p>∴ Answer: A</p>
Correct Answer: A