Let the sets 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 = . The maximum number of such possible partitions of S is equal to

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Solution

Q.
Let the sets $S = {2, 4, 8, 16, \dots, 512}$ be partitioned into 3 sets $A, B, C$ with equal number of elements such that $A \cup B \cup C = S$ and $A \cap B = B \cap C = A \cap C = ϕ .$ The maximum number of such possible partitions of $S$ is equal to [2024]

1 1520

2 1640

3 1710

4 1680

Ans.
(4)

We have, $S = {2, 4, 8, 16, \dots, 512}$

$= {2^{1}, 2^{2}, 2^{3}, 2^{4}, \dots, 2^{9}}$ so, $| S | = 9$

$| A | = | B | = | C |$ and $A \cup B \cup C = S$

and $A \cap B = B \cap C = A \cap C = ϕ$

Since, $| S | = 9$ so $| A | = | B | = | C | = 3$

So, number of ways of making partition of $S = {}^{9}C_{3} \times {}^{6}{C_{3}} \times {}^{3}{C_{3}}$

$= \frac{9!}{6! 3!} \times \frac{6!}{3! \times 3!} \times 1$

$= \frac{9!}{3! \times 3! \times 3!} = 1680$

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