Let S=\1,2,3,4,5,6\ . Then the number of one-one functions f: S arrow P(S) , where P(S) denote the power set of S , such that f(n) \subset f(~m) where n \lt m is \underline\hspace2cm .

Functions PYQ — JEE Main 2023 · 30 Jan · Shift 1 · Q37 with Solution

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Correct answer: 3240

$f:S\to P(S),\qquad S=\{1,2,3,4,5,6\}$

such that for all $n<m$ ,

$f(n)\subset f(m).$

So the images form a strictly increasing chain:

$f(1)\subset f(2)\subset f(3)\subset f(4)\subset f(5)\subset f(6).$

Since each inclusion is strict, injectivity is automatically satisfied.

Let

$A_i=f(i),\qquad i=1,2,\dots,6.$

Then

$A_1\subset A_2\subset A_3\subset A_4\subset A_5\subset A_6\subseteq S.$

Define

\quad B_2=A_2\setminus A_1, \quad B_3=A_3\setminus A_2, \quad \dots, \quad B_6=A_6\setminus A_5, \quad B_7=S\setminus A_6.$$ Then: - $B_1,B_2,\dots,B_7$ are pairwise disjoint, - their union is $S$, - because each inclusion $A_i\subset A_{i+1}$ is strict, we must have $$B_2,B_3,B_4,B_5,B_6\neq \varnothing.$$ But $B_1$ and $B_7$ may be empty or nonempty. Thus counting such functions is equivalent to counting ways to distribute the 6 elements of $S$ into 7 labeled boxes $$B_1,B_2,B_3,B_4,B_5,B_6,B_7$$ with boxes $2,3,4,5,6$ nonempty. --- 3. Use inclusion-exclusion / surjection counting. First place each of the 6 elements into one of the 7 labeled boxes: total $$7^6.$$ We require 5 specific boxes ($B_2$ to $B_6$) to be nonempty. By inclusion-exclusion, the count is $$\sum_{j=0}^{5}(-1)^j\binom{5}{j}(7-j)^6.$$ Compute: $$\binom{5}{0}7^6=117649$$ $$\binom{5}{1}6^6=5\cdot 46656=233280$$ $$\binom{5}{2}5^6=10\cdot 15625=156250$$ $$\binom{5}{3}4^6=10\cdot 4096=40960$$ $$\binom{5}{4}3^6=5\cdot 729=3645$$ $$\binom{5}{5}2^6=64$$ So $$117649-233280+156250-40960+3645-64.$$ Now simplify: $$117649-233280=-115631$$ $$-115631+156250=40619$$ $$40619-40960=-341$$ $$-341+3645=3304$$ $$3304-64=3240.$$ Hence the number of such functions is $$\boxed{3240}.$$ --- 4. Comparison with stored answer Derived answer = $3240$. Stored correct answer = $3240$. They match.