Let the set S={2,4,8,16, …, 512} be partitioned into 3 sets A, B, C with equal number of elements such that A ∪ B ∪ C=S and A ∩ B=B ∩ C=A ∩ C=φ. The maximum number of such possible partitions of S is equal to:

Solution

Maths sir · 4 steps

Nine powers of 2 go into three named boxes of 3 each, so pick A's three, then B's, and C gets the rest.

1. Count the set. S = {2¹, 2², …, 2⁹} because 512 = 2⁹. So S has 9 elements, and each of A, B, C gets exactly 3.
2. Fill A, then B. The sets are named, so putting {2,4,8} in A is different from putting it in B. Choose 3 of the 9 for A in C(9, 3) = 84 ways, then 3 of the remaining 6 for B in C(6, 3) = 20 ways.
3. C is forced. The last 3 elements must go to C, which is C(3, 3) = 1 way. The three sets are disjoint and their union is S, as required.

   C(9, 3)C(6, 3)C(3, 3) = 84 × 20 × 1 = 1680 = 9!/(3! 3! 3!)

4. So. The number of partitions is 1680, which is option (d).

The idea to keep. Distributing n distinct items into named groups of sizes p, q, r gives n!/(p! q! r!). Divide by 3! only when the equal-sized groups have no names.

Answer from the NTA final answer key. Solution by PrepWiser.