Explanation:

Part-1

\( |A| = (d-e) - c(a-b) \)

Since \( : \ a, b, c, d, e \in \{0, 1\} \)

Total possibilities \( = 32 \)

Part-2

Only integral values taken by \( |A| \) can be \( \pm 1 \)

\( \Rightarrow \) We want \( |A| \) to be odd

So, bad cases are those with \( |A| \) even

Let’s count bad cases.

Case : \( 1 \) \( (d-e) \) and \( c(a-b) \) are even

For \( d-e \) to be even \( : d = e \Rightarrow 2 \) cases

Part-3

For \( c(a-b) \) to be even

\( c = 0 \Rightarrow 4 \) cases

\( c = 1 \Rightarrow a = b \Rightarrow 2 \) case

Total \( 12 \) cases

Case-2: \( d-e \) and \( c(a-b) \) are odd

\( d-e \) is odd \( \Rightarrow d \ne e \Rightarrow 2 \) cases

\( c(a-b) \) is odd \( \Rightarrow c = 1 \) and \( a \ne b \Rightarrow 2 \) cases

Total \( 4 \) cases

\( \Rightarrow \) Bad cases \( = 16 \)

And Good cases \( = 16 \) too

ALTERNATIVELY : There is a bijective between good and bad cases. We can switch the value of \( d \) to map a good case to bad case. But we want \( |A| \in \{-1, 1\} \).

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