If T denotes the number of ordered triplets $(A_{1}, A_{2}, A_{3})$ of sets which have the property that:(i) $A_{1} \cup A_{2} \cup A_{3} = {1, 2, 3, ...., 10, 11}$ and (ii) $A_{1} \cap A_{2} \cap A_{3} = ϕ$ then which of the following options(s) is/are true?

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Number of positive integral divisors of T is 144

Number of ways T can be written as product of two positive integers (order doesn’t matter) such that their H.C.F. to be 3 is 2

Number of positive integral divisors of T of the form $(3 K + 1)$ (where $K = 0, 1, 2, 3, . . . . .$ ) is 6

Sum of positive integral divisors of T is $(3^{12} - 1) (2^{12} - 1)$

answer is A, B, C.

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Detailed Solution

$T = 6^{11} = 2^{11} \times 3^{11}$

$∴$ Number of divisors of T is $= 12 \times 12 = 144$
If we need H.C.F. to be 3 then let $3^{11}$ is divided as 3 and $3^{10}$ .
Now $2^{11}$ will go completely with 3 or $3^{10}$

$∴$ Number of ways =2
Number of divisors of T of the form $3 K + 1$ , we do not need any 3.
Now, we can see odd powers of 2 are of the form $3 K + 2$ and even powers are of the form $3 K + 1$ which can be done in 6 ways.
Sum of divisors of $T = (2^{0} + 2^{1} + ..... + 2^{11}) (3^{0} + 3^{1} + ..... + 3^{11})$
$= \frac{(2^{12} - 1)}{2 - 1} . \frac{(3^{12} - 1)}{3 - 1} = \frac{1}{2} (2^{12} - 1) (3^{12} - 1)$