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Exactly 2 elements in S will point to 1 element in R. Therefore, total combinations=> mC2 * nC

Rest (m-2) elements in S will point to exactly 1 of (n-1) elements in R. So, no of 1:1 functions => (n-1)P(m-2)

Multiplying both results, we get =>

n! m(m-1) / 2* (n-m+1)!

So the answer must be B. There is a typo in that option.

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