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Consider two sets A={2,3,5,7,11,13} and B={1,8,27}. Let f be a function from A to B such that for every element b in B, there is at least one element a in A such that f(a)=b. Then, the total number of such functions f is

CAT · 2024 · Quant Slot 1
Question:
Consider two sets A={2,3,5,7,11,13} and B={1,8,27}. Let f be a function from A to B such that for every element b in B, there is at least one element a in A such that f(a)=b. Then, the total number of such functions f is

Options

540
537
665
667
Answer: 540

Explanation:
Step 1: Formula for surjective functions Number of surjective functions from a set of size m to a set of size n: n! × S(m, n), where S(m, n) is the Stirling number of the second kind (number of ways to partition m elements into n non-empty subsets). Here: m = 6, n = 3 Step 2: Find S(6, 3) Formula for Stirling numbers: S(m, n) = (1/n!) × ∑_{k=0}^{n} (−1)^k × C(n, k) × (n − k)^m S(6,3) = (1/3!) × [C(3,0)*3^6 − C(3,1)*2^6 + C(3,2)*1^6 − C(3,3)*0^6] Compute step by step: 3^6 = 729 2^6 = 64 1^6 = 1 0^6 = 0 Sum = 1729 − 364 + 31 − 10 = 729 − 192 + 3 = 540 Divide by 3! = 6 → S(6,3) = 540 / 6 = 90 Step 3: Total surjective functions Total = n! × S(6,3) = 3! × 90 = 6 × 90 = 540

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