JEE Mathematics 2023 Question with Solution

Given: $A = \{1,2,3,4,5\}$, $B = \{1,2,3,4,5,6\}$, and the function $f : A \to B$ satisfies $f(1) + f(2) = f(4) - 1$.

Find: The total number of such functions.

From

$$f(1) + f(2) = f(4) - 1$$

we get

$$f(4) = f(1) + f(2) + 1$$

Since $f(4) \in B = \{1,2,3,4,5,6\}$, we must have

$$f(1) + f(2) \le 5$$

Now count ordered pairs $\bigl(f(1), f(2)\bigr)$ with values in $\{1,2,3,4,5,6\}$ such that their sum is at most $5$.

• If $f(1) = 1$, then $f(2) = 1,2,3,4$: $4$ choices.
• If $f(1) = 2$, then $f(2) = 1,2,3$: $3$ choices.
• If $f(1) = 3$, then $f(2) = 1,2$: $2$ choices.
• If $f(1) = 4$, then $f(2) = 1$: $1$ choice.

Hence the number of possible pairs $\bigl(f(1), f(2)\bigr)$ is

$$4 + 3 + 2 + 1 = 10$$

For each such pair, $f(4)$ is uniquely determined by

$$f(4) = f(1) + f(2) + 1$$

Also, $f(3)$ and $f(5)$ can each be chosen freely from $B$, so each has $6$ choices.

Therefore, the total number of functions is

$$10 \times 6 \times 6 = 360$$

Thus, the required number of functions is $360$.

Given: The condition is $f(1) + f(2) = f(4) - 1$.

Find: How many functions $f : A \to B$ satisfy it.

Rewrite the condition as

$$f(4) = f(1) + f(2) + 1$$

Since $f(4) \in \{1,2,3,4,5,6\}$, the sum $f(1) + f(2)$ can only be $1,2,3,4,5$. But because both $f(1)$ and $f(2)$ are at least $1$, the possible sums are actually $2,3,4,5$.

Count all ordered pairs:

$$\begin{aligned} f(1)+f(2)=2 &\Rightarrow 1 \text{ pair} \\ f(1)+f(2)=3 &\Rightarrow 2 \text{ pairs} \\ f(1)+f(2)=4 &\Rightarrow 3 \text{ pairs} \\ f(1)+f(2)=5 &\Rightarrow 4 \text{ pairs} \end{aligned}$$

So total valid ordered pairs are

$$1+2+3+4=10$$

For every valid pair $\bigl(f(1), f(2)\bigr)$, the value of $f(4)$ is fixed.

The remaining unconstrained function values are $f(3)$ and $f(5)$, each having $6$ choices. Hence,

$$\text{Total functions} = 10 \cdot 6 \cdot 6 = 360$$

Therefore, the answer is $360$.

Given: $f(4) = f(1) + f(2) + 1$.

Find: The total number of valid functions.

A quick way is to count valid ordered pairs $\bigl(f(1), f(2)\bigr)$ by using the condition $f(1)+f(2) \le 5$. The number of positive integer ordered pairs with sum at most $5$ is

$$1+2+3+4 = 10$$

For each such pair, $f(4)$ is determined uniquely. The remaining free choices are $f(3)$ and $f(5)$, giving

$$6^2 = 36$$

choices. Thus,

$$10 \times 36 = 360$$

So the required number of functions is $360$.