JEE Main 2023 — Functions Question with Solution

Given that the function $$f:A\to B$$ 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 $$f:A\to B$$ satisfying the condition $$f(1)+f(2)=f(4)-1$$ is 360.