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Solution


Q.

Let A = {1, 3, 7, 9, 11} and B = {2, 4, 5, 7, 8, 10, 12}. Then the total number of one-one maps f:A→B, such that f(1)+f(3)=14, is:           [2024]
1 180  
2 480  
3 120  
4 240  

Ans.

(4)

   A = {1, 3, 7, 9, 11}, B = {2, 4, 5, 7, 8, 10, 12}

   f:A→B is one-one such that

   f(1)+f(3)=14

   (f(1),f(3))={(2,12),(4,10),(12,2),(10,4)}

   Since f is one-one so f(1)≠f(3) so we cannot take (7, 7).

   ∴  So, for f(1) we have 4 choices and for f(3) we have 4 choices and remaining 3 elements have 5! choices for mapping to be one-one.

   Total number of ways = 4×5!/2

   =4802=240                                  [∵ Pair (2, 12) and (12, 2) will be considered same]