If (f:A\to B) is a function and (A={p,q,r,s}), (B={0,1}), how many functions are possible with (f(p)=0) and (f(q)=1)?

Correct answer: (4)

The direct answer is B: 4 functions are possible. A function from \(A\) to \(B\) gives one value in \(B=\{0,1\}\) to every element of \(A\). The values of \(p\) and \(q\) are already fixed: \(f(p)=0\) and \(f(q)=1\). Only \(r\) and \(s\) remain free. Each of these two inputs independently has 2 choices, 0 or 1. Therefore the number is \(2\times2=2^2=4\). Option A, 2, counts only one choice for each remaining input and misses the combinations. Option B, 4, is correct. Option C, 8, would correspond to three free inputs, but only two are free. Option D, 16, is the unrestricted count \(2^4\) when none of the four values is fixed; it ignores the two conditions. The four possibilities for \((f(r),f(s))\) are \((0,0),(0,1),(1,0),(1,1)\). Memory cue: fixed inputs contribute no choice; multiply the number of codomain choices for every unfixed input.