JEE Mathematics 2024 Question with Solution

Given: $A = \{1,2,3,4,5,6,7\}$, so $|A| = 7$. The function is $f: A \to P(A)$ with the condition $a \in f(a)$ for every $a \in A$.

Find: The value of $m+n$ if the number of such functions is $m^n$ with least possible $m$.

For a fixed element $a \in A$, the subset $f(a)$ must contain $a$. Hence, among the $7$ elements of $A$, one element is forced to be present and the remaining $6$ elements can be chosen freely.

So the number of possible choices for each $f(a)$ is

$$2^6 = 64$$

Since this choice is independent for each of the $7$ elements of the domain, the total number of such functions is

$$(2^6)^7 = 2^{42}$$

Now write this as $m^n$ with least possible base. Since $2^{42}$ already has the smallest natural base greater than $1$, we get

$$m = 2, \quad n = 42$$

Therefore,

$$m+n = 2+42 = 44$$

So, the required answer is $44$.

Given: Each element of $A$ must belong to its own image subset.

Find: $m+n$.

For every input element, one member is compulsory in the image subset, and the other $6$ members are optional. So each input has exactly

$$2^6$$

possible images.

With $7$ independent inputs, total functions are

$$(2^6)^7 = 2^{42}$$

Thus the least base is $m=2$ and $n=42$, giving

$$m+n=44$$

Hence, the answer is $44$.