Can be the empty set ?

Yes, because includes the empty set , and holds for any non-empty subset .

Why are and allowed to be empty?

can be empty because . Similarly, does not have to be , so can be non-empty or empty.

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Permutations and Combinations: Mathematics | JEE Main

Q: Why do all four intermediate cases give 360?

In each intermediate case, 5 steps remove 1 element and 1 step removes 2 elements. The number of ways is because choosing the sequence of removals is equivalent to partitioning 6 elements into blocks of size (1, 1, 1, 1, 2).

Let

S = \{1,2,3,4,5,6\}

. Then the number of one-one functions

f: S \to P(S)

, where

P(S)

denote the power set of

S

, such that

f(n) \subset f(m)

where

n < m

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Step-by-step solutionView

Correct answer

The total number of such one-one functions is 3240.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

$S = \{1, 2, 3, 4, 5, 6\}$ with $|S| = 6$ . We seek the number of one-one functions $f: S \to P(S)$ such that $f(1) \subset f(2) \subset f(3) \subset f(4) \subset f(5) \subset f(6)$ .

02goal

Find the total number of strictly increasing chains of subsets $A_1 \subset A_2 \subset A_3 \subset A_4 \subset A_5 \subset A_6 \subseteq S$ .

03approach

Let $k_i = |f(i)|$ . Since the inclusions are strict, we have $0 \le k_1 < k_2 < k_3 < k_4 < k_5 < k_6 \le 6$ . There are 6 integers in the range $[0, 6]$ which has 7 possible values. Thus, exactly one size from $\{0, 1, 2, 3, 4, 5, 6\}$ is omitted in the sequence of cardinalities.

04execute

Case 1: The omitted size is 0 or 1. That is, $k_6=6, k_5=5, k_4=4, k_3=3, k_2=2$ , and $k_1 \in \{0, 1\}$ . - Number of ways to choose $f(6) = S$ is 1. - Number of ways to choose $f(5) \subset f(6)$ with $|f(5)|=5$ is $\binom{6}{5} = 6$ . - Number of ways to choose $f(4) \subset f(5)$ with $|f(4)|=4$ is $\binom{5}{4} = 5$ . - Number of ways to choose $f(3) \subset f(4)$ with $|f(3)|=3$ is $\binom{4}{3} = 4$ . - Number of ways to choose $f(2) \subset f(3)$ with $|f(2)|=2$ is $\binom{3}{2} = 3$ . - For $f(1)$ , it can be either $\emptyset$ (1 option) or any 1-element subset of $f(2)$ ( $\binom{2}{1} = 2$ options). Thus there are $1 + 2 = 3$ choices. Total for Case 1: $1 \times 6 \times 5 \times 4 \times 3 \times 3 = 1080$ .

05execute

Case 2: The omitted size is $j \in \{2, 3, 4, 5\}$ . Here $k_1 = 0$ (so $f(1) = \emptyset$ , 1 choice) and $k_6 = 6$ (so $f(6) = S$ , 1 choice). At the omitted size $j$ , the step size in cardinality is 2 instead of 1 (a jump of 2 elements). For any such omitted size $j \in \{2, 3, 4, 5\}$ , the number of choices is: - When size 5 is omitted: $\binom{6}{4} \times \binom{4}{3} \times \binom{3}{2} \times \binom{2}{1} \times 1 = 15 \times 4 \times 3 \times 2 \times 1 = 360$ . - By symmetry of element drops, for each omitted size in $\{2, 3, 4, 5\}$ , the number of chains is 360. Total for these 4 cases: $4 \times 360 = 1440$ .

06execute

Case 3: The omitted size is 6. Thus the sizes are $(0, 1, 2, 3, 4, 5)$ . $f(6)$ is any 5-element subset of $S$ : $\binom{6}{5} = 6$ ways. Then sizes decrease by 1 down to 0: $\binom{6}{5} \times \binom{5}{4} \times \binom{4}{3} \times \binom{3}{2} \times \binom{2}{1} \times 1 = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$ .

07execute

Summing all mutually exclusive cases: $1080 + 1440 + 720 = 3240$ .

✓verify

Alternative partition: A strictly increasing chain of 6 subsets of a 6-element set corresponds to placing each of the 6 elements of $S$ into one of the differences $A_1$ , $A_2 \setminus A_1$ , ..., $A_6 \setminus A_5$ , $S \setminus A_6$ . There are 7 regions, exactly one of which has size 2 (or two have size 0, etc.). All 7 choices for the omitted cardinality are exhausted and verified.

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Can

f(1)

be the empty set

\emptyset

Yes, because $P(S)$ includes the empty set $\emptyset$ , and $\emptyset \subset f(2)$ holds for any non-empty subset $f(2)$ .

Why do all four intermediate cases give 360?

In each intermediate case, 5 steps remove 1 element and 1 step removes 2 elements. The number of ways is $6! / 2! = 720 / 2 = 360$ because choosing the sequence of removals is equivalent to partitioning 6 elements into blocks of size (1, 1, 1, 1, 2).

Why are

D_1

and

D_7

allowed to be empty?

$f(1) = D_1$ can be empty because $\emptyset \in P(S)$ . Similarly, $f(6)$ does not have to be $S$ , so $S \setminus f(6) = D_7$ can be non-empty or empty.

Answer

The total number of such one-one functions is 3240.

Why each option works or fails

Step-by-step solution

given: $S = \{1, 2, 3, 4, 5, 6\}$ with $|S| = 6$ . We seek the number of one-one functions $f: S \to P(S)$ such that $f(1) \subset f(2) \subset f(3) \subset f(4) \subset f(5) \subset f(6)$ .
goal: Find the total number of strictly increasing chains of subsets $A_1 \subset A_2 \subset A_3 \subset A_4 \subset A_5 \subset A_6 \subseteq S$ .
approach: Let $k_i = |f(i)|$ . Since the inclusions are strict, we have $0 \le k_1 < k_2 < k_3 < k_4 < k_5 < k_6 \le 6$ . There are 6 integers in the range $[0, 6]$ which has 7 possible values. Thus, exactly one size from $\{0, 1, 2, 3, 4, 5, 6\}$ is omitted in the sequence of cardinalities.
execute: Case 1: The omitted size is 0 or 1. That is, $k_6=6, k_5=5, k_4=4, k_3=3, k_2=2$ , and $k_1 \in \{0, 1\}$ . - Number of ways to choose $f(6) = S$ is 1. - Number of ways to choose $f(5) \subset f(6)$ with $|f(5)|=5$ is $\binom{6}{5} = 6$ . - Number of ways to choose $f(4) \subset f(5)$ with $|f(4)|=4$ is $\binom{5}{4} = 5$ . - Number of ways to choose $f(3) \subset f(4)$ with $|f(3)|=3$ is $\binom{4}{3} = 4$ . - Number of ways to choose $f(2) \subset f(3)$ with $|f(2)|=2$ is $\binom{3}{2} = 3$ . - For $f(1)$ , it can be either $\emptyset$ (1 option) or any 1-element subset of $f(2)$ ( $\binom{2}{1} = 2$ options). Thus there are $1 + 2 = 3$ choices. Total for Case 1: $1 \times 6 \times 5 \times 4 \times 3 \times 3 = 1080$ .
execute: Case 2: The omitted size is $j \in \{2, 3, 4, 5\}$ . Here $k_1 = 0$ (so $f(1) = \emptyset$ , 1 choice) and $k_6 = 6$ (so $f(6) = S$ , 1 choice). At the omitted size $j$ , the step size in cardinality is 2 instead of 1 (a jump of 2 elements). For any such omitted size $j \in \{2, 3, 4, 5\}$ , the number of choices is: - When size 5 is omitted: $\binom{6}{4} \times \binom{4}{3} \times \binom{3}{2} \times \binom{2}{1} \times 1 = 15 \times 4 \times 3 \times 2 \times 1 = 360$ . - By symmetry of element drops, for each omitted size in $\{2, 3, 4, 5\}$ , the number of chains is 360. Total for these 4 cases: $4 \times 360 = 1440$ .
execute: Case 3: The omitted size is 6. Thus the sizes are $(0, 1, 2, 3, 4, 5)$ . $f(6)$ is any 5-element subset of $S$ : $\binom{6}{5} = 6$ ways. Then sizes decrease by 1 down to 0: $\binom{6}{5} \times \binom{5}{4} \times \binom{4}{3} \times \binom{3}{2} \times \binom{2}{1} \times 1 = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$ .
execute: Summing all mutually exclusive cases: $1080 + 1440 + 720 = 3240$ .
verify: Alternative partition: A strictly increasing chain of 6 subsets of a 6-element set corresponds to placing each of the 6 elements of $S$ into one of the differences $A_1$ , $A_2 \setminus A_1$ , ..., $A_6 \setminus A_5$ , $S \setminus A_6$ . There are 7 regions, exactly one of which has size 2 (or two have size 0, etc.). All 7 choices for the omitted cardinality are exhausted and verified.

Shortcut:

When to use it: Fastest conceptual method: viewing the nested chain as distributing the 6 elements into 7 distinct intervals/differences.

given: Chain $\emptyset \subseteq A_1 \subset A_2 \subset A_3 \subset A_4 \subset A_5 \subset A_6 \subseteq S$ with $|S|=6$ .

execute: Let $d_i = |D_i|$ . We have $d_1 + d_2 + d_3 + d_4 + d_5 + d_6 + d_7 = 6$ with $d_2, d_3, d_4, d_5, d_6 \ge 1$ and $d_1, d_7 \ge 0$ . Since $\sum_{i=2}^6 d_i \ge 5$ , the sum of these 5 positive integers can be 5 or 6. Case A: $\sum_{i=2}^6 d_i = 5$ , so $d_2 = d_3 = d_4 = d_5 = d_6 = 1$ . Then $d_1 + d_7 = 1$ . There are 2 choices for which is 1 and which is 0. For each choice, the sizes of $(D_1, \dots, D_7)$ are six 1s and one 0. The number of ways to distribute the 6 elements is $6! = 720$ . Thus, $2 \times 720 = 1440$ ways. Case B: $\sum_{i=2}^6 d_i = 6$ , so $d_1 = d_7 = 0$ . Exactly one $d_i$ ( $i \in \{2, 3, 4, 5, 6\}$ ) is 2 and the other four are 1. There are 5 choices for which $d_i = 2$ . The number of ways to distribute the 6 elements is $5 \times \frac{6!}{2! 1! 1! 1! 1!} = 5 \times 360 = 1800$ ways. Total $= 1440 + 1800 = 3240$ .

verify: Both cases are disjoint and exhaust all non-negative integer solutions to $\sum_{i=1}^7 d_i = 6$ with $d_2, \dots, d_6 \ge 1$ : $1440 + 1800 = 3240$ .

Permutations and Combinations: Mathematics | JEE Main

What feels right?

The total number of such one-one functions is 3240.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

07execute

✓verify

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