QUESTION 157
The number of ways to arrange 4 people in a row is
(A) 12
(B) 16
(C) 20
(D) 24
(E) 28

To stimulate his daughter's reasoning, a father made the following drawing and gave it to the child along with three colored pencils. He wants the girl to paint only the circles, so that those connected by a segment have different colors.
In how many different ways can the child do what the father asked?
(A) 6
(B) 12
(C) 18
(D) 24
(E) 72

How many ways can 4 people be arranged in a line?
(A) 8
(B) 12
(C) 16
(D) 20
(E) 24

There are $n$ students in a class and no two of them have the same height. The students stand in a line, one behind another, in no particular order of their heights.
(a) How many different orders are there in which the shortest student is not in the first position and the tallest student is not in the last position?
(b) The badness of an ordering is the largest number $k$ with the following property. There is at least one student $X$ such that there are $k$ students taller than $X$ standing ahead of $X$. Find a formula for $g _ { k } ( n ) =$ number of orderings of $n$ students with badness $k$.
Example: The ordering $64\,61\,67\,63\,62\,66\,65$ (the numbers denote heights) has badness 3 as the student with height 62 has three taller students (with heights 64, 67 and 63) standing ahead in the line and nobody has more than 3 taller students standing ahead.
Possible hints for (b): It may be useful to first count orderings of badness 1 and/or to find $f _ { k } ( n ) =$ the number of orderings of $n$ students with badness less than or equal to $k$.

Two adults and three children go to an amusement park to ride a certain ride. This ride has 2 chairs in the front row and 3 chairs in the back row. When each child must sit in the same row as an adult, find the number of ways for all 5 people to sit in the chairs of the ride. [4 points]

How many ways are there to select 3 letters from the four letters $a , b , c , d$ with repetition allowed and arrange them in a row? [2 points]
(1) 56
(2) 60
(3) 64
(4) 68
(5) 72

6. The number of ways to arrange 3 people in a row is
A. $12$ ways
B. $18$ ways
C. $24$ ways
D. $36$ ways [Figure] [Figure] [Figure] [Figure] [Figure]
C. C and D can know each other's scores
B. B can know all four people's scores

10. Given that the foci of ellipse $C$ are $F _ { 1 } ( - 1,0 ) , F _ { 2 } ( 1,0 )$, and a line through $F _ { 2 }$ intersects $C$ at points $A , B$. If $\left| A F _ { 2 } \right| = 2 \left| F _ { 2 } B \right| , | A B | = \left| B F _ { 1 } \right|$, then the equation of $C$ is
A. $\frac { x ^ { 2 } } { 2 } + y ^ { 2 } = 1$
B. $\frac { x ^ { 2 } } { 3 } + \frac { y ^ { 2 } } { 2 } = 1$
C. $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 3 } = 1$
D. $\frac { x ^ { 2 } } { 5 } + \frac { y ^ { 2 } } { 4 } = 1$

In how many ways can $10$ A's and $6$ B's be arranged in a row such that no two B's are adjacent and no two A's are adjacent? (More precisely: find the number of arrangements of $10$ A's and $6$ B's such that between any two consecutive B's there is at least one A, and between any two consecutive A's there is at least one B.)

Consider all the permutations of the twenty six English letters that start with $z$. In how many of these permutations the number of letters between $z$ and $y$ is less than those between $y$ and $x$?
(A) $6 \times 23!$
(B) $6 \times 24!$
(C) $156 \times 23!$
(D) $156 \times 24!$.

Suppose Roger has 4 identical green tennis balls and 5 identical red tennis balls. In how many ways can Roger arrange these 9 balls in a line so that no two green balls are next to each other and no three red balls are together?
(A) 8
(B) 9
(C) 11
(D) 12

Let $S_n$ be the set of all $n$-digit numbers whose digits are all 1 or 2 and there are no consecutive 2's. (Example: 112 is in $S_3$ but 221 is not in $S_3$). Then the number of elements in $S_{10}$ is
(A) 512
(B) 256
(C) 144
(D) 89

Let $n$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let $m$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then the value of $\frac { m } { n }$ is

Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated ?
(1) $2 ! 3 ! 4$ !
(2) $( 3 ! ) ^ { 3 } \cdot ( 4 ! )$
(3) $( 3 ! ) 2 . ( 4 ! )$
(4) $3 ! ( 4 ! ) ^ { 3 }$

The number of 9-digit numbers, that can be formed using all the digits of the number 123456789, such that the even digits occupy only even places, is
(1) 2880
(2) 2520
(3) 2160
(4) 2400

The number of permutations, of the digits $1, 2, 3, \ldots, 7$ without repetition, which neither contain the string 153 nor the string 2467, is $\_\_\_\_$.

In a group of 3 girls and 4 boys, there are two boys $B _ { 1 }$ and $B _ { 2 }$. The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but $B _ { 1 }$ and $B _ { 2 }$ are not adjacent to each other, is :
(1) 96
(2) 144
(3) 120
(4) 72

The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is

There is a staircase of 10 steps which we are to climb. We can go up one step at a time or two steps at a time, but we have to use each method at least once.
(1) Suppose we can go up two steps at a time twice or more in a row. (i) If we climb the staircase going up two steps at a time just 3 times, we will go up one step at a time just $\mathbf{P}$ times, and there are $\mathbf{QR}$ different ways of climbing the staircase. (ii) If we can go up two steps at a time twice or more in a row, there are altogether $\mathbf{ST}$ different ways of climbing the staircase.
(2) Suppose we cannot go up two steps at a time twice or more in a row. (i) If we climb the staircase going up two steps at a time just twice, we will go up one step at a time just $\mathbf{U}$ times, and there are $\mathbf{VW}$ different ways of climbing the staircase. (ii) If we cannot go up two steps at a time twice or more in a row, there are altogether $\mathbf{XY}$ different ways of climbing the staircase.

There is a staircase of 10 steps which we are to climb. We can go up one step at a time or two steps at a time, but we have to use each method at least once.
(1) Suppose we can go up two steps at a time twice or more in a row. (i) If we climb the staircase going up two steps at a time just 3 times, we will go up one step at a time just $\mathbf{P}$ times, and there are $\mathbf{QR}$ different ways of climbing the staircase. (ii) If we can go up two steps at a time twice or more in a row, there are altogether $\mathbf{ST}$ different ways of climbing the staircase.
(2) Suppose we cannot go up two steps at a time twice or more in a row. (i) If we climb the staircase going up two steps at a time just twice, we will go up one step at a time just $\mathbf{U}$ times, and there are $\mathbf{VW}$ different ways of climbing the staircase. (ii) If we cannot go up two steps at a time twice or more in a row, there are altogether $\mathbf{XY}$ different ways of climbing the staircase.

There are six students (three female and three male) who frequently interact with a teacher at school. After graduation, the teacher invites them to a gathering. After the meal, seven people stand in a row for a commemorative photo. It is known that among the students, one female and one male had an unpleasant experience and do not want to stand adjacent during the photo, while the teacher stands in the middle and the three male students do not all stand on the same side of the teacher. The total number of possible arrangements is (17--1)(17--2)(17--3).

Arrange the digits $1, 2, 3, \ldots, 9$ into a nine-digit number (digits cannot be repeated) such that the first 5 digits are increasing from left to right and the last 5 digits are decreasing from left to right. How many nine-digit numbers satisfy the conditions?
(1) $\frac{8!}{4!4!}$
(2) $\frac{8!}{5!3!}$
(3) $\frac{9!}{5!4!}$
(4) $\frac{8!}{5!}$
(5) $\frac{9!}{5!}$

A school is holding a concert with 5 piano performances, 4 violin performances, and 3 vocal performances, totaling 12 different pieces. The school wants to arrange performances of the same type together, and vocal performances must come after either piano or violin performances. How many possible arrangements of pieces are there for this concert?
(1) $5 ! \times 4 ! \times 3 !$
(2) $2 \times 5 ! \times 4 ! \times 3 !$
(3) $3 \times 5 ! \times 4 ! \times 3 !$
(4) $4 \times 5 ! \times 4 ! \times 3 !$
(5) $6 \times 5 ! \times 4 ! \times 3 !$

Three domestic automobiles of brands A, B, and C and three foreign automobiles of brands X, Y, and Z will be displayed in a single row at an exhibition according to the following conditions.

• Domestic and foreign automobiles will be arranged consecutively within their own groups.
• Brand A automobile will be in the first or last position among all automobiles.
• Brand X automobile will be in the first or last position among foreign automobiles.

Given this, in how many different ways can the automobiles be displayed?
A) 10
B) 12
C) 14
D) 16
E) 18

Two students from each of three different schools will participate in a chess tournament. In the first round of the tournament, each student will be paired with a student who is not from their own school.
Accordingly, in how many different ways can the pairing in the first round be done?
A) 6
B) 8
C) 9
D) 12
E) 15