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Step-by-Step Solution
Step 1: Understand the Problem
We have two sets:
A = \{1, 2, 3, 4, 5\} and
B = \{1, 2, 3, 4, 5, 6\} .
We want to find the number of functions
f: A \rightarrow B
that satisfy the condition
f(1) + f(2) = f(4) - 1.
Rearranging the given condition:
f(1) + f(2) + 1 = f(4) .
Each of f(1), f(2), f(3), f(4), f(5) can only take values from B = \{1, 2, 3, 4, 5, 6\}.
Step 2: Determine Possible Values for f(4)
From f(1) + f(2) + 1 = f(4) , the left side must be at least 1 + 1 + 1 = 3 and at most 6 + 6 + 1 = 13 . However, f(4) can only be from \{1, 2, 3, 4, 5, 6\} . Thus, feasible values for f(4) given the condition are 3, 4, 5, or 6 .
Step 3: Count Valid Pairs (f(1), f(2)) for Each f(4)
Case (a): f(4) = 6
Equation: f(1) + f(2) + 1 = 6 \implies f(1) + f(2) = 5.
Possible pairs (f(1), f(2)) from B that sum to 5: (1,4), (2,3), (3,2), (4,1). (4 possibilities)
For each valid pair, f(3) and f(5) can independently be any of the 6 values in B .
Total functions for this case: 4 \times 6 \times 6 = 144.
Case (b): f(4) = 5
Equation: f(1) + f(2) + 1 = 5 \implies f(1) + f(2) = 4.
Possible pairs (f(1), f(2)) that sum to 4: (1,3), (2,2), (3,1). (3 possibilities)
For each valid pair, f(3) and f(5) can be any of the 6 values in B .
Total functions for this case: 3 \times 6 \times 6 = 108.
Case (c): f(4) = 4
Equation: f(1) + f(2) + 1 = 4 \implies f(1) + f(2) = 3.
Possible pairs (f(1), f(2)) that sum to 3: (1,2), (2,1). (2 possibilities)
For each valid pair, f(3) and f(5) can be any of the 6 values in B .
Total functions for this case: 2 \times 6 \times 6 = 72.
Case (d): f(4) = 3
Equation: f(1) + f(2) + 1 = 3 \implies f(1) + f(2) = 2.
Possible pair (f(1), f(2)) that sums to 2: (1,1). (1 possibility)
For this single valid pair, f(3) and f(5) can each be any of the 6 values in B .
Total functions for this case: 1 \times 6 \times 6 = 36.
Step 4: Sum Up All Valid Cases
Adding up the results from all four cases:
144 + 108 + 72 + 36 = 360.
Final Answer
The total number of functions satisfying the condition is 360 .
