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 :
- 120
- 180
- 240
- 480
Answer
(3) 240
Solution
Pairs of distinct elements of B with sum 14: \2, 12\, \4, 10\ (7 + 7 not allowed, f is one-one)
∵ each pair can be given to (f(1), f(3)) in 2 orders 2 2 = 4 ways
Remaining 3 elements of A → remaining 5 elements of B one-one: 5 4 3 = 60 ways
Total = 4 60 = 240
Answer: (3)