If the domain and range of f(x) = ^{9-x}C_{x- | vedclass

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(B) For the function $f(x) = ^{9-x}C_{x-1}$ to be defined,the following conditions must be satisfied: $1$. $9-x \geq 0 \Rightarrow x \leq 9$ $2$. $x-1

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$m = n + 1$

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(B) For the function $f(x) = ^{9-x}C_{x-1}$ to be defined,the following conditions must be satisfied: $1$. $9-x \geq 0 \Rightarrow x \leq 9$ $2$. $x-1 \geq 0 \Rightarrow x \geq 1$ $3$. $9-x \geq x-1 \Rightarrow 10 \geq 2x \Rightarrow x \leq 5$ Combining these,we get $1 \leq x \leq 5$. Since $x$ must be an integer for the combination to be defined,$x \in \{1, 2, 3, 4, 5\}$. Thus,the domain has $m = 5$ elements. Now,we calculate the values of $f(x)$ for each $x$: $f(1) = ^{9-1}C_{1-1} = ^8C_0 = 1$ $f(2) = ^{9-2}C_{2-1} = ^7C_1 = 7$ $f(3) = ^{9-3}C_{3-1} = ^6C_2 = 15$ $f(4) = ^{9-4}C_{4-1} = ^5C_3 = 10$ $f(5) = ^{9-5}C_{5-1} = ^4C_4 = 1$ The range is the set of distinct values: $\{1, 7, 15, 10\}$. Thus,the range has $n = 4$ elements. Comparing $m=5$ and $n=4$,we see that $m = n + 1$.

If the domain and range of $f(x) = ^{9-x}C_{x-1}$ contain $m$ and $n$ elements respectively,then:

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