IPMAT Indore 2022 · Quantitative Ability · Question 12
IPMAT Indore 2022 previous year question on Set Theory, with the answer and a step-by-step solution.
Answer & solution
Answer: 16
The rule. For a finite set X with \left| X \right|=k, the power set has
\left| P\left( X \right) \right|={2}^{k}
Build up one layer at a time, starting from A=⌀ with \left| A \right|=0:
Set | Size |
|---|---|
A | 0 |
P\left( A \right) | {2}^{0}=1 |
P\left( P\left( A \right) \right) | {2}^{1}=2 |
P\left( P\left( P\left( A \right) \right) \right) | {2}^{2}=4 |
P\left( P\left( P\left( P\left( A \right) \right) \right) \right) | {2}^{4}=16 |
Worth seeing concretely at the start: P\left( ⌀ \right)=\{⌀\} — the empty set has exactly one subset, namely itself — so the first layer has size 1, not 0. That single step is where the question is won or lost.
Answer: 16
Shortcut. The sizes go 0\to 1\to 2\to 4\to 16, each being 2 raised to the previous one. Note the jump at the end: {2}^{4}=16, not {2}^{3}=8 — the exponent is the previous size, which was already 4.
The one trap is the first step. \left| P\left( ⌀ \right) \right|=1, because ⌀ is a subset of itself. Starting from 0 and writing {2}^{0}=1 handles it automatically; starting by “there are no subsets” gives 0 and derails everything after.
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