Let \(S=\{1,2,3,4,5,6\}\). Then the number of one-one functions \(f: S \rightarrow P(S)\), where \(P(S)\) denote the power set of \(S\), such that \(f(n) \subset f(m)\) where \(\mathrm{n}<\mathrm{m}\) is ______.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Sets Relations and Functions

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

is ______.

Answer: 3240

A one-one

f

with

f (n) \subset f (m)

for

n < m

is a strictly increasing chain of six subsets of the six-element set, so their sizes are six distinct values chosen from

{0, 1, 2, 3, 4, 5, 6}

— exactly one size is skipped. For a chain of sizes

s_{1} < s_{2} < \dots < s_{6}

the number of such nested selections is

\frac{6 !}{s _{1} ! ( s _{2} - s _{1} )! \dots ( s _{6} - s _{5} )! ( 6 - s _{6} )!} .

Skipping size

0

or size

6

leaves all gaps equal to

1

, giving

\frac{6 !}{1} = 720

each; skipping any one of the five interior sizes creates a single gap of

2

, giving

\frac{6 !}{2 !} = 360

each. The total is

2 (720) + 5 (360) = 1440 + 1800 = 3240.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Integer answer