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(D) Let $S$ be the set of all strictly increasing functions from $A = \{1, 2, 3, 4, 5, 6\}$ to $B = \{1, 2, 3, \dots, 9\}$. The total number of such f

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(D) Let $S$ be the set of all strictly increasing functions from $A = \{1, 2, 3, 4, 5, 6\}$ to $B = \{1, 2, 3, \dots, 9\}$. The total number of such functions is $\binom{9}{6} = \binom{9}{3} = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.We want to find the number of functions such that $f(i) 
eq i$ for all $i \in \{1, 2, 3, 4, 5, 6\}$.Let $P_i$ be the property that $f(i) = i$. We want to find the number of functions satisfying none of the properties $P_1, P_2, \dots, P_6$.Using the Principle of Inclusion-Exclusion, the number of such functions is $\sum_{k=0}^{6} (-1)^k S_k$, where $S_k$ is the sum of the number of functions satisfying at least $k$ specific properties.For a strictly increasing function, if $f(i_1) = i_1, f(i_2) = i_2, \dots, f(i_k) = i_k$ with $i_1  i_k$, $f(j) > j$. Since $f$ is strictly increasing, this forces $f(x) = x$ for all $x \in \{1, \dots, 6\}$.However, a simpler approach is to count directly: Let $f(1) = x_1, f(2) = x_2, \dots, f(6) = x_6$ where $1 \le x_1 Total strictly increasing functions = $\binom{9}{6} = 84$.Using the complement: Total - (functions where at least one $f(i)=i$).By calculating the valid cases: The number of strictly increasing functions $f: \{1, \dots, n\} 	o \{1, \dots, m\}$ such that $f(i) 
eq i$ is given by the formula $\binom{m-1}{n}$.Here $n=6, m=9$, so the number is $\binom{9-1}{6} = \binom{8}{6} = \binom{8}{2} = \frac{8 	imes 7}{2} = 28$.

The number of strictly increasing functions $f$ from the set $A = \{1, 2, 3, 4, 5, 6\}$ to the set $B = \{1, 2, 3, \dots, 9\}$ such that $f(i) \neq i$ for all $1 \le i \le 6$ is equal to:

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