JEE MAIN - Mathematics (2023 - 11th April Evening Shift - No. 16)
Let $$\mathrm{A}=\{1,2,3,4,5\}$$ and $$\mathrm{B}=\{1,2,3,4,5,6\}$$. Then the number of functions $$f: \mathrm{A} \rightarrow \mathrm{B}$$ satisfying $$f(1)+f(2)=f(4)-1$$ is equal to __________.
Answer
360
Explanation
Given that the function $$f : A \rightarrow B$$ satisfies the condition $$f(1) + f(2) = f(4) - 1$$, where the set $$A = \{1, 2, 3, 4, 5\}$$ and the set $$B = \{1, 2, 3, 4, 5, 6\}$$.
We want to find out how many such functions exist.
First, observe that the condition $$f(1) + f(2) = f(4) - 1$$ can be rewritten as $$f(1) + f(2) + 1 = f(4)$$. So, the sum of $$f(1), f(2),$$ and 1 is equal to $$f(4)$$. Since $$f(4)$$ is a value in set B, it can take values from 1 to 6.
The maximum value of $$f(1) + f(2) + 1$$ can be $$6 + 6 + 1 = 13$$, but this is more than 6 (the maximum value of $$f(4)$$), so it's not possible. Thus, the maximum value of $$f(4)$$ in this case can be 6.
Let's now analyze the number of functions for each value of $$f(4)$$ from 3 to 6 (we start from 3 because $$f(1)$$ and $$f(2)$$ take values from set B and their minimum sum plus 1 is 3):
1. When $$f(4) = 6$$, then $$f(1) + f(2) = 5$$. The pairs $$(f(1), f(2))$$ that satisfy this equation are $$(1, 4), (2, 3), (3, 2), (4, 1)$$. For each of these 4 cases, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$4 \times 6 \times 6 = 144$$ functions.
2. When $$f(4) = 5$$, then $$f(1) + f(2) = 4$$. The pairs $$(f(1), f(2))$$ that satisfy this equation are $$(1, 3), (2, 2), (3, 1)$$. For each of these 3 cases, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$3 \times 6 \times 6 = 108$$ functions.
3. When $$f(4) = 4$$, then $$f(1) + f(2) = 3$$. The pairs $$(f(1), f(2))$$ that satisfy this equation are $$(1, 2), (2, 1)$$. For each of these 2 cases, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$2 \times 6 \times 6 = 72$$ functions.
4. When $$f(4) = 3$$, then $$f(1) + f(2) = 2$$. The only pair $$(f(1), f(2))$$ that satisfies this equation is $$(1, 1)$$. For this case, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$1 \times 6 \times 6 = 36$$ functions.
Adding the numbers of functions from all these cases, we get a total of $$144 + 108 + 72 + 36 = 360$$ functions from $$A$$ to $$B$$ that satisfy the given condition.
Therefore, the number of functions $$f : A \rightarrow B$$ satisfying the condition $$f(1) + f(2) = f(4) - 1$$ is 360.
We want to find out how many such functions exist.
First, observe that the condition $$f(1) + f(2) = f(4) - 1$$ can be rewritten as $$f(1) + f(2) + 1 = f(4)$$. So, the sum of $$f(1), f(2),$$ and 1 is equal to $$f(4)$$. Since $$f(4)$$ is a value in set B, it can take values from 1 to 6.
The maximum value of $$f(1) + f(2) + 1$$ can be $$6 + 6 + 1 = 13$$, but this is more than 6 (the maximum value of $$f(4)$$), so it's not possible. Thus, the maximum value of $$f(4)$$ in this case can be 6.
Let's now analyze the number of functions for each value of $$f(4)$$ from 3 to 6 (we start from 3 because $$f(1)$$ and $$f(2)$$ take values from set B and their minimum sum plus 1 is 3):
1. When $$f(4) = 6$$, then $$f(1) + f(2) = 5$$. The pairs $$(f(1), f(2))$$ that satisfy this equation are $$(1, 4), (2, 3), (3, 2), (4, 1)$$. For each of these 4 cases, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$4 \times 6 \times 6 = 144$$ functions.
2. When $$f(4) = 5$$, then $$f(1) + f(2) = 4$$. The pairs $$(f(1), f(2))$$ that satisfy this equation are $$(1, 3), (2, 2), (3, 1)$$. For each of these 3 cases, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$3 \times 6 \times 6 = 108$$ functions.
3. When $$f(4) = 4$$, then $$f(1) + f(2) = 3$$. The pairs $$(f(1), f(2))$$ that satisfy this equation are $$(1, 2), (2, 1)$$. For each of these 2 cases, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$2 \times 6 \times 6 = 72$$ functions.
4. When $$f(4) = 3$$, then $$f(1) + f(2) = 2$$. The only pair $$(f(1), f(2))$$ that satisfies this equation is $$(1, 1)$$. For this case, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$1 \times 6 \times 6 = 36$$ functions.
Adding the numbers of functions from all these cases, we get a total of $$144 + 108 + 72 + 36 = 360$$ functions from $$A$$ to $$B$$ that satisfy the given condition.
Therefore, the number of functions $$f : A \rightarrow B$$ satisfying the condition $$f(1) + f(2) = f(4) - 1$$ is 360.
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