The following is a process for finding the number of functions $f : X \rightarrow X$ where $X = \{ 1,2,3,4,5,6 \}$ such that the range of the composite function $f \circ f$ has 5 elements. Let the ranges of functions $f$ and $f \circ f$ be $A$ and $B$, respectively. If $n ( A ) = 6$, then $f$ is a bijection, and $f \circ f$ is also a bijection, so $n ( B ) = 6$. Also, if $n ( A ) \leq 4$, then $B \subset A$, so $n ( B ) \leq 4$. Therefore, we only need to consider the case where $n ( A ) = 5$, that is, $B = A$.
(i) The number of ways to choose a subset $A$ of $X$ with $n ( A ) = 5$ is (가).
(ii) For the set $A$ chosen in (i), let $k$ be the element of $X$ that does not belong to $A$. Since $n ( A ) = 5$, the number of ways to choose $f ( k )$ from set $A$ is (나).
(iii) For $A = \left\{ a _ { 1 } , a _ { 2 } , a _ { 3 } , a _ { 4 } , a _ { 5 } \right\}$ chosen in (i) and $f ( k )$ chosen in (ii), since $f ( k ) \in A$ and $A = B$, we have $A = \left\{ f \left( a _ { 1 } \right) , f \left( a _ { 2 } \right) , f \left( a _ { 3 } \right) , f \left( a _ { 4 } \right) , f \left( a _ { 5 } \right) \right\} \cdots ( * )$. The number of cases satisfying (*) is equal to the number of bijections from set $A$ to set $A$, so it is (다). Therefore, by (i), (ii), and (iii), the number of functions $f$ we seek is (가) $\times$ (나) $\times$ (다). When the numbers corresponding to (가), (나), and (다) are $p$, $q$, and $r$, respectively, what is the value of $p + q + r$? [4 points]
(1) 131
(2) 136
(3) 141
(4) 146
(5) 151

The following is a process to find the number of functions $f$ such that the number of elements in the range of the composite function $f \circ f$ is 5, for the set $X = \{ 1,2,3,4,5,6 \}$ and the function $f : X \rightarrow X$.
Let the ranges of the function $f$ and the function $f \circ f$ be $A$ and $B$, respectively. If $n ( A ) = 6$, then $f$ is a bijection, and $f \circ f$ is also a bijection, so $n ( B ) = 6$. Also, if $n ( A ) \leq 4$, then $B \subset A$, so $n ( B ) \leq 4$. Therefore, we only need to consider the case where $n ( A ) = 5$, that is, $B = A$.
(i) The number of ways to choose a subset $A$ of $X$ with $n ( A ) = 5$ is (가).
(ii) For the set $A$ chosen in (i), let $k$ be the element of $X$ that does not belong to $A$. Since $n ( A ) = 5$, the number of ways to choose $f ( k )$ from the set $A$ is (나).
(iii) For $A = \left\{ a _ { 1 } , a _ { 2 } , a _ { 3 } , a _ { 4 } , a _ { 5 } \right\}$ chosen in (i) and $f ( k )$ chosen in (ii), since $f ( k ) \in A$ and $A = B$, we have $A = \left\{ f \left( a _ { 1 } \right) , f \left( a _ { 2 } \right) , f \left( a _ { 3 } \right) , f \left( a _ { 4 } \right) , f \left( a _ { 5 } \right) \right\} \cdots ( * )$. The number of cases satisfying (*) is equal to the number of bijections from set $A$ to set $A$, so $\square$ (다).
Therefore, by (i), (ii), and (iii), the number of functions $f$ we seek is $\square$ (가) $\times$ $\square$ (나) $\times$ $\square$ (다).
When the numbers that fit (가), (나), and (다) are $p , q , r$ respectively, what is the value of $p + q + r$? [4 points]
(1) 131
(2) 136
(3) 141
(4) 146
(5) 151

For two sets $X = \{ 1,2,3,4,5 \} , Y = \{ 1,2,3,4 \}$, how many functions $f$ from $X$ to $Y$ satisfy the following conditions? [4 points]
(a) For all elements $x$ in set $X$, $f ( x ) \geq \sqrt { x }$.
(b) The range of function $f$ has exactly 3 elements.
(1) 128
(2) 138
(3) 148
(4) 158
(5) 168

For every pair $(p, k)$ of nonzero natural numbers, we denote by $S(p, k)$ the number of surjections from $\llbracket 1, p \rrbracket$ to $\llbracket 1, k \rrbracket$. Consistently, for every $p \in \mathbb{N}^*$, we set $S(p, 0) = 0$.
Determine $S(n+1, n)$.

For every pair $(p, k)$ of nonzero natural numbers, we denote by $S(p, k)$ the number of surjections from $\llbracket 1, p \rrbracket$ to $\llbracket 1, k \rrbracket$. Consistently, for every $p \in \mathbb{N}^*$, we set $S(p, 0) = 0$.
How many applications are there from $\llbracket 1, p \rrbracket$ to $\llbracket 1, n \rrbracket$?

For every pair $(p, k)$ of nonzero natural numbers, we denote by $S(p, k)$ the number of surjections from $\llbracket 1, p \rrbracket$ to $\llbracket 1, k \rrbracket$. Consistently, for every $p \in \mathbb{N}^*$, we set $S(p, 0) = 0$.
For $p \geqslant n$, establish the formula $$n^p = \sum_{k=0}^{n} \binom{n}{k} S(p, k)$$ where $S(p, 0) = 0$ by convention.

For every pair $(p, k)$ of nonzero natural numbers, we denote by $S(p, k)$ the number of surjections from $\llbracket 1, p \rrbracket$ to $\llbracket 1, k \rrbracket$. Consistently, for every $p \in \mathbb{N}^*$, we set $S(p, 0) = 0$.
Deduce an expression of $S(p, n)$ for $p \geqslant n$.

For every pair $(p, k)$ of nonzero natural numbers, we denote by $S(p, k)$ the number of surjections from $\llbracket 1, p \rrbracket$ to $\llbracket 1, k \rrbracket$. Consistently, for every $p \in \mathbb{N}^*$, we set $S(p, 0) = 0$.
By rereading question I.B.5, comment on the consistency of the expression of $S(p,n)$ for $p < n$.

For every pair $(p, k)$ of nonzero natural numbers, we denote by $S(p, k)$ the number of surjections from $\llbracket 1, p \rrbracket$ to $\llbracket 1, k \rrbracket$. Consistently, for every $p \in \mathbb{N}^*$, we set $S(p, 0) = 0$.
Simplify as much as possible the following expressions: $$\sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} k^n \quad \text{and} \quad \sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} k^{n+1}$$

The number of maps $f$ from the set $\{ 1, 2, 3 \}$ into the set $\{ 1, 2, 3, 4, 5 \}$ such that $f(i) \leq f(j)$ whenever $i < j$ is
(A) 60
(B) 50
(C) 35
(D) 30

Consider three boxes, each containing 10 balls labelled $1, 2, \ldots, 10$. Suppose one ball is drawn from each of the boxes. Denote by $n_i$, the label of the ball drawn from the $i$-th box, $i = 1, 2, 3$. Then the number of ways in which the balls can be chosen such that $n_1 < n_2 < n_3$ is
(A) 120
(B) 130
(C) 150
(D) 160

The number of maps $f$ from the set $\{ 1,2,3 \}$ into the set $\{ 1,2,3,4,5 \}$ such that $f ( i ) \leq f ( j )$ whenever $i < j$ is
(a) 60
(b) 50
(c) 35
(d) 30

The number of maps $f$ from the set $\{ 1,2,3 \}$ into the set $\{ 1,2,3,4,5 \}$ such that $f ( i ) \leq f ( j )$ whenever $i < j$ is
(a) 60
(b) 50
(c) 35
(d) 30

The number of maps $f$ from the set $\{ 1, 2, 3 \}$ into the set $\{ 1, 2, 3, 4, 5 \}$ such that $f(i) \leq f(j)$ whenever $i < j$ is
(A) 60
(B) 50
(C) 35
(D) 30

Consider three boxes, each containing 10 balls labelled $1, 2, \ldots, 10$. Suppose one ball is drawn from each of the boxes. Denote by $n_i$, the label of the ball drawn from the $i$-th box, $i = 1, 2, 3$. Then the number of ways in which the balls can be chosen such that $n_1 < n_2 < n_3$ is
(A) 120
(B) 130
(C) 150
(D) 160

The number of maps $f$ from the set $\{ 1, 2, 3 \}$ into the set $\{ 1, 2, 3, 4, 5 \}$ such that $f ( i ) \leq f ( j )$ whenever $i < j$ is
(A) 60
(B) 50
(C) 35
(D) 30

Consider three boxes, each containing 10 balls labelled $1, 2, \ldots, 10$. Suppose one ball is drawn from each of the boxes. Denote by $n _ { i }$, the label of the ball drawn from the $i$-th box, $i = 1, 2, 3$. Then the number of ways in which the balls can be chosen such that $n _ { 1 } < n _ { 2 } < n _ { 3 }$ is
(A) 120
(B) 130
(C) 150
(D) 160

Let $A = \{1,2,3,4,5,6\}$ and $B = \{a,b,c,d,e\}$. How many functions $f : A \rightarrow B$ are there such that for every $x \in A$, there is one and exactly one $y \in A$ with $y \neq x$ and $f(x) = f(y)$?
(A) 450
(B) 540
(C) 900
(D) 5400.

Let $A = \left\{ x _ { 1 } , x _ { 2 } , \ldots , x _ { 50 } \right\}$ and $B = \left\{ y _ { 1 } , y _ { 2 } , \ldots , y _ { 20 } \right\}$ be two sets of real numbers. What is the total number of functions $f : A \rightarrow B$ such that $f$ is onto and $f \left( x _ { 1 } \right) \leq f \left( x _ { 2 } \right) \leq \cdots \leq f \left( x _ { 50 } \right)$ ?
(A) $\binom { 49 } { 19 }$
(B) $\binom { 49 } { 20 }$
(C) $\binom { 50 } { 19 }$
(D) $\binom { 50 } { 20 }$

Let $( n _ { 1 } , n _ { 2 } , \cdots , n _ { 12 } )$ be a permutation of the numbers $1,2 , \cdots , 12$. The number of arrangements with $$n _ { 1 } > n _ { 2 } > n _ { 3 } > n _ { 4 } > n _ { 5 } > n _ { 6 }$$ and $$n _ { 6 } < n _ { 7 } < n _ { 8 } < n _ { 9 } < n _ { 10 } < n _ { 11 } < n _ { 12 }$$ equals:
(A) $\binom { 12 } { 5 }$
(B) $\binom { 12 } { 6 }$
(C) $\binom { 11 } { 6 }$
(D) $\frac { 11 ! } { 2 }$

How many functions $f : \{ 1,2 , \ldots , 10 \} \rightarrow \{ 1 , \ldots , 2000 \}$, which satisfy $$f ( i + 1 ) - f ( i ) \geq 20 , \text { for all } 1 \leq i \leq 9 ,$$ are there?
(A) $10 ! \binom { 1829 } { 10 }$
(B) $11 ! \binom { 1830 } { 11 }$
(C) $\binom { 1829 } { 10 }$
(D) $\binom { 1830 } { 11 }$

Let $X$ be a set with exactly 5 elements and $Y$ be a set with exactly 7 elements. If $\alpha$ is the number of one-one functions from $X$ to $Y$ and $\beta$ is the number of onto functions from $Y$ to $X$, then the value of $\frac { 1 } { 5 ! } ( \beta - \alpha )$ is $\_\_\_\_$ .

Statement 1: If $A$ and $B$ be two sets having $p$ and $q$ elements respectively, where $q > p$. Then the total number of functions from set $A$ to set $B$ is $q^{p}$. Statement 2: The total number of selections of $p$ different objects out of $q$ objects is ${}^{q}C_{p}$.
(1) Statement 1 is true, Statement 2 is false.
(2) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.
(3) Statement 1 is false, Statement 2 is true.
(4) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1.

The number of bijective functions $f:\{1,3,5,7,\cdots,99\} \rightarrow \{2,4,6,8,\cdots,100\}$ if $f(3) > f(5) > f(7) \cdots > f(99)$ is
(1) ${}^{50}C_1$
(2) ${}^{50}C_2$
(3) $\frac{50!}{2}$
(4) ${}^{50}C_3 \times 3!$

Let $A = \{ ( x , y ) : 2 x + 3 y = 23 , x , y \in \mathbb { N } \}$ and $B = \{ x : ( x , y ) \in A \}$. Then the number of one-one functions from $A$ to $B$ is equal to $\_\_\_\_$