JEE Mathematics 2023 Question with Solution

Given: $S = \{1,2,3,4,5,6\}$ so $n(S)=6$.

Find: The number of one-one functions $f:S \to P(S)$ such that $f(n) \subset f(m)$ whenever $$n

$$$$

Case 1

$f(6)=S$ : $1$ option.

$f(5)$ = any $5$-element subset of $S$ : $6$ options.

$f(4)$ = any $4$-element subset of $f(5)$ : $5$ options.

$f(3)$ = any $3$-element subset of $f(4)$ : $4$ options.

$f(2)$ = any $2$-element subset of $f(3)$ : $3$ options.

$f(1)$ = any $1$-element subset of $f(2)$ or the empty subset : $3$ options.

Hence,

$$1 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 3 = 1080$$

Case 2

$f(6)$ = any $5$-element subset of $S$ : $6$ options.

$f(5)$ = any $4$-element subset of $f(6)$ : $5$ options.

$f(4)$ = any $3$-element subset of $f(5)$ : $4$ options.

$f(3)$ = any $2$-element subset of $f(4)$ : $3$ options.

$f(2)$ = any $1$-element subset of $f(3)$ : $2$ options.

$f(1)$ = the empty subset : $1$ option.

So,

$$6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 720$$

Case 3

$f(6)=S$ : $1$ option.

$f(5)$ = any $4$-element subset of $S$ : $15$ options.

$f(4)$ = any $3$-element subset of $f(5)$ : $4$ options.

$f(3)$ = any $2$-element subset of $f(4)$ : $3$ options.

$f(2)$ = any $1$-element subset of $f(3)$ : $2$ options.

$f(1)$ = the empty subset : $1$ option.

Therefore,

$$1 \cdot 15 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 360$$

Cases 4, 5 and 6

Similarly, the remaining configurations contribute $360$ functions each.

So the total is

$$1080 + 720 + 360 + 360 + 360 + 360 = 3240$$

Therefore, the number of such functions is $3240$.