Question 10

I realize since 2^x values are all distinct, g(f(x))=2^x forces f to be injective, and any injective f works by defining g appropriately. So I need to count injective functions from a 5-element set to a 7-element set with the constraints f(2)≠2 and f(4)≠4, starting from the total count of 7·6·5·4·3 = 2520. Using inclusion-exclusion: fixing f(2)=2 gives 6·5·4·3=360, same for f(4)=4, and fixing both gives 5·4·3=60. So the answer is 2520 − 720 + 60 = 1860. **Key observation:** Since the values $2^x$ for $x\in A$ are all distinct, the condition $g(f(x)) = 2^x$ forces $f$ to be injective (if $f(x_1)=f(x_2)$ with $x_1\ne x_2$, then $2^{x_1}=2^{x_2}$, contradiction). Conversely, if $f$ is injective, define $g(f(x)) = 2^x$ and $g$ arbitrarily elsewhere. So $T$ = set of injective $f:A\to B$ with $f(2)\neq 2,\ f(4)\neq 4$. - Total injective maps: $7\cdot6\cdot5\cdot4\cdot3 = 2520$ - With $f(2)=2$: $6\cdot5\cdot4\cdot3 = 360$; with $f(4)=4$: $360$ - Both: $5\cdot4\cdot3 = 60$ By inclusion–exclusion: $2520 - 360 - 360 + 60 = 1860$.

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For g to exist, f must be injective, since g(f(x))=2^x forces distinct inputs to map to distinct f-values. So T consists of injective functions A→B avoiding f(2)=2 and f(4)=4, and I need to count these, starting with the total injective count P(7,5)=7·6·5·4·3. That gives 2520 total injective functions. Using inclusion-exclusion: fixing f(2)=2 leaves 360 ways for the rest, same for f(4)=4, and fixing both gives 60. So the answer is 2520 - 360 - 360 + 60 = 1860. **Reasoning:** For $g\circ f(x)=2^x$ to be well-defined as a function of $x$, whenever $f(x_1)=f(x_2)$ we'd need $2^{x_1}=2^{x_2}$, forcing $x_1=x_2$. So $f$ must be **injective**, and conversely any injective $f$ allows defining $g$ appropriately (assign $g(f(x))=2^x$ on the image, and arbitrary values elsewhere in $B$). So $T$ = injective functions $f:A\to B$ with $f(2)\neq 2$, $f(4)\neq 4$. Total injective functions $A\to B$: $P(7,5)=7\cdot6\cdot5\cdot4\cdot3=2520$. Let $X$: injective with $f(2)=2$. Fixing $f(2)=2$, remaining 4 elements map injectively into remaining 6 values: $P(6,4)=360$. Let $Y$: injective with $f(4)=4$: similarly $360$. $X\cap Y$: $f(2)=2, f(4)=4$ fixed, remaining 3 elements into remaining 5 values: $P(5,3)=60$. By inclusion-exclusion: $$|T| = 2520 - 360 - 360 + 60 = 1860.$$ ```json