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## Chosing Program | AMC 10A, 2013 | Problem 7

Try this beautiful problem from Combinatorics: Chosing Program

## Chosing Program – AMC-10A, 2013- Problem 7

A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?

,

• $6$
• $8$
• $9$
• $12$
• $16$

### Key Concepts

Combinatorics

Answer: $9$

AMC-10A (2013) Problem 7

Pre College Mathematics

## Try with Hints

There are six programms: English, Algebra, Geometry, History, Art, and Latin. Since the student must choose a program of four course with the condition that there must contain English and at least one mathematics course. Therefore one course( i.e English) are already fixed and we have to find out the other subjects combinations…….

Can you now finish the problem ……….

There are Two cases :
Case 1: The student chooses both algebra and geometry.
This means that 3 courses have already been chosen. We have 3 more options for the last course, so there are 3 possibilities here.
case 2: The student chooses one or the other.
Here, we simply count how many ways we can do one, multiply by 2 , and then add to the previous.

Let us choose the mathematics course is algebra. so we can choose 2 of History, Art, and Latin, which is simply $3 \choose 2$=$3$. If it is geometry, we have another 3 options, so we have a total of 6 options if only one mathematics course is chosen.

can you finish the problem……..

Therefore the require ways are $6+3=9$

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## Probability in Divisibility | AMC-10A, 2003 | Problem 15

Try this beautiful problem from AMC 10A, 2003 based on Probability in Divisibility.

## Probability in Divisibility – AMC-10A, 2003- Problem 15

What is the probability that an integer in the set ${1,2,3,…,100}$ is divisible by $2$ and not divisible by $3$?

• $\frac {33}{100}$
• $\frac{1}{6}$
• $\frac{17}{50}$
• $\frac{1}{2}$
• $\frac{18}{25}$

### Key Concepts

Number system

Probability

divisibility

Answer: $\frac{17}{50}$

AMC-10A (2003) Problem 15

Pre College Mathematics

## Try with Hints

There are total number of integers are $100$.and numer of integers divisible by $2$ is $\frac{100}{2}$=$50$. Now we have to find out divisible by $2$ and not divisible by $3$. so at first we have to find out the numbers of integers which are divisible by $2$ and $3$ both…….

can you finish the problem……..

To be divisible by both $2$ and $3$, a number must be divisible by the lcm of $(2,3)=6$.

Therefore numbers of integers which are divisible by $6$=$\frac{100}{6}=16$ (between $1$ & $100$)

can you finish the problem……..

Therefore the number of integers which are divisible by $2$ and not divisible by $3$= $50 – 16=34$.

So require probability= $\frac{34}{100}=\frac{17}{50}$

Categories

## Probability in Game | AMC-10A, 2005 | Problem 18

Try this beautiful problem from AMC 10A, 2005 based on Probability in Game.

## Probability in Game – AMC-10A, 2005- Problem 18

Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?

• $\frac{1}{4}$
• $\frac{1}{6}$
• $\frac{1}{5}$
• $\frac{2}{3}$
• $\frac{1}{3}$

### Key Concepts

Probability

combinatorics

Answer: $\frac{1}{5}$

AMC-10A (2005) Problem 18

Pre College Mathematics

## Try with Hints

Given that  The first team to win three games wins the series, team B wins the second game and team A wins the series. So the Total number of games played=$5$. Now we have to find out the possible order of wins…..

Can you now finish the problem ……….

Possible cases :

If team B won the first two games, team A would need to win the next three games. Therefore the possible order of wins is BBAAA.
If team A won the first game, and team B won the second game, the possible order of wins is $A B B A A, A B A B A,$ and $A B A A X,$ where $X$ denotes that the 5th game wasn’t played.
since ABAAX is dependent on the outcome of 4 games instead of 5, it is twice as likely to occur and can be treated as two possibilities.

According to the question, there is One possibility where team $\mathrm{B}$ wins the first game and 5 total possibilities, Therefore the required probability is $\frac{ 1}{5}$

Categories

## Fair coin Problem | AIME I, 1990 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on fair coin.

## Fair Coin Problem – AIME I, 1990

A fair coin is to be tossed 10 times. Let i|j, in lowest terms, be the probability that heads never occur on consecutive tosses, find i+j.

• is 107
• is 73
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Combinatorics

Algebra

AIME I, 1990, Question 9

Elementary Algebra by Hall and Knight

## Try with Hints

5 tails flipped, any less,

by Pigeonhole principle there will be heads that appear on consecutive tosses

(H)T(H)T(H)T(H)T(H)T(H) 5 tails occur there are 6 slots for the heads to be placed but only 5H remaining, ${6 \choose 5}$ possible combination of 6 heads there are

$\sum_{i=6}^{11}{i \choose 11-i}$=${6 \choose 5} +{7 \choose 4}+{8 \choose 3}+{9 \choose 2} +{10 \choose 1} +{11 \choose 0}$=144

there are $2^{10}$ possible flips of 10 coins

or, probability=$\frac{144}{1024}=\frac{9}{64}$ or, 9+64=73.

Categories

## Head Tail Problem | AIME I, 1986 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1986 based on Head Tail Problem.

## Head Tail Problem – AIME I, 1986

In a sequence of coin tosses, one can keep a record of instances in which a tail is immediately followed by a head , a head is immediately followed by ahead and etc. We denote these by TH, HH, and etc. For example in the sequence TTTHHTHTTTHHTTH of 15 coin tosses we observe that there are two HH, three HT, four TH and five TT subsequences. How many different sequences of 15 coin tosses will contain exactly two HH, three HT, four TH, and five TT subsequences?

• is 107
• is 560
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Combinatorics

Algebra

AIME I, 1986, Question 13

Elementary Algebra by Hall and Knight

## Try with Hints

Let us observe the sequences.

H switches to T three times, T switches to H four times.

There are 5 TT subsequences.

We are to add 5 T’s into, the string. There are already 4 T’s in the sequence.

We are to add 5 balls in 4 urns which is same as 3 dividers ${5+3 \choose 3}$=56

We do the same with 2H’s to get ${2+3 \choose 3}$=10

so, $56 \times 10$=560.

Categories

## Combinatorics in Tournament | AIME I, 1985 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1985 based on Combinatorics in Tournament.

## Combinatorics in Tournament- AIME I, 1985

In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded 1 point, the loser got 0 points, and each of the two player earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?

• is 107
• is 25
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Combinatorics

Algebra

AIME I, 1985, Question 14

Elementary Algebra by Hall and Knight

## Try with Hints

Let there be n+10 players

Case I from n players not in weakest 10, ${n \choose 2}$ games played and ${n \choose 2}$ points earned

Case II n players also earned ${n \choose 2}$ points against weakest 10

Case III now weakest 10 played among themselves ${10 \choose 2}$=45 games and 45 points earned

Case IV 10 players also earned 45 points against stronger n

So total points earned= 2[${n \choose 2}$+45]=$n^{2}-n+90$

case V 1 point earned per game ${n+10 \choose 2}$=$\frac{(n+10)(n+9)}{2}$ games and $\frac{(n+10)(n+9)}{2}$ points earned

So $n^{2}-n+90=\frac{(n+10)(n+9)}{2}$

or, $n^{2}-21n+90=0$

or, n=6, n=15 here taking n>10,

or, n=15 or, n+10=25.

Categories

## Regular polygon | Combinatorics | PRMO-2019 | Problem 15

Try this beautiful problem from combinatorics PRMO 2019 based on Regular polygon

## Regular polygon| PRMO | Problem 15

In how many ways can a pair of parallel diagonals of a regular polygon of $10$ sides be selected

• $24$
• $45$
• $34$

### Key Concepts

Combinatorics

Regular polygon

geometry

Answer:$45$

PRMO-2019, Problem 15

Pre College Mathematics

## Try with Hints

The above diagram is a diagram of Regular Polygon .we have to draw the diagonals as shown in above.we joined the diagonals such that all the diagonals will be parallel

Can you now finish the problem ……….

If we joined the diagonals (shown in Fig. 1), i.e $(P_3 \to P_10)$,$(P_4\to P_9)$,$(P_5 \to P_8)$ then then we have 3 diagonals.so we have 5$4 \choose 2$ ways=$15$ ways.

If we joined the diagonals (shown in Fig.2), i.e $(P_1 \to P_3)$,$(P_10\to P_4)$,$(P_9\to P_5)$,$(P_8\to P_6)$then we have $4$diagonals.so we have 5$3 \choose 2$ ways=$30$ ways.

Therefore total numbers of ways that can a pair of parallel diagonals of a regular polygon of $10$ sides be selected is $15+30=45$

Categories

## Row of Pascal Triangle | AIME I, 1992 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Row of Pascal Triangle.

## Row of Pascal Triangle – AIME I, 1992

In Pascal’s Triangle, each entry is the sum of the two entries above it. Find the row of Pascal’s triangle do three consecutive entries occur that are in the ratio 3:4:5.

• is 107
• is 62
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Digits

Combinatorics

AIME I, 1992, Question 4

Elementary Number Theory by David Burton

## Try with Hints

For consecutive entries

$\frac{{n \choose (x-1)}}{3}=\frac{{n \choose x}}{4}=\frac{{n \choose {x+1}}}{5}$

from first two terms $\frac{n!}{3(x-1)!(n-x+1)!}=\frac{n!}{4x!(n-x)!}$

$\Rightarrow \frac{1}{3(n-x+1)}=\frac{1}{4x}$

$\Rightarrow \frac{3(n+1)}{7}=x$ is first equation

for the next two terms

$\frac{n!}{4(n-x)!x!}=\frac{n!}{5(n-x-1)!(x+1)!}$

$\Rightarrow \frac{4(n-x)}{5}=x+1$

$\Rightarrow \frac{4n}{5}=\frac{9x}{5}+1$

from first equation putting value of x here gives

$\Rightarrow \frac{4n}{5}=\frac{9 \times 3(n+1)}{5 \times 7}+1$

$\Rightarrow n=62, x=\frac{3(62+1)}{7}=27$

$\Rightarrow$ n=62.

Categories

## Quadratic equation Problem | AMC-10A, 2002 | Problem 12

Try this beautiful problem from Algebra based Quadratic equation.

## Quadratic equation Problem – AMC-10A, 2002- Problem 12

Both roots of the quadratic equation $x^2 – 63x + k = 0$ are prime numbers. The number of possible values of $k$ is

• $0$
• $1$
• $2$
• $4$
• more than $4$

### Key Concepts

Algebra

prime numbers

Answer: $1$

AMC-10A (2002) Problem 12

Pre College Mathematics

## Try with Hints

The given equation is $x^2 – 63x + k = 0$. Say that the roots are primes…

Comparing the equation with $ax^2 +bx+c=0$ we get $a=1 , b=-63 , c=k$.. Let $m_1$ & $m_2$ be the roots of the given equation…

using vieta’s Formula we may sat that…$m_1 + m_2 =-(- 63)=63$ and $m_1 m_2 = k$

can you finish the problem……..

Now the roots are prime. Sum of the two roots are $63$ and product is $k$

Therefore one root must be $2$ ,otherwise the sum would be even number

can you finish the problem……..

So other root will be $63-2$=$61$. Therefore product must be $m_1m_2=122$

Hence the answer is $1$

Categories

## Largest possible value | AMC-10A, 2004 | Problem 15

Try this beautiful problem from Number system: largest possible value

## Largest Possible Value – AMC-10A, 2004- Problem 15

Given that $-4 \leq x \leq -2$ and $2 \leq y \leq 4$, what is the largest possible value of $\frac{x+y}{2}$

• $\frac {-1}{2}$
• $\frac{1}{6}$
• $\frac{1}{2}$
• $\frac{1}{4}$
• $\frac{1}{9}$

### Key Concepts

Number system

Inequality

divisibility

Answer: $\frac{1}{2}$

AMC-10A (2003) Problem 15

Pre College Mathematics

## Try with Hints

The given expression is $\frac{x+y}{x}=1+\frac{y}{x}$

Now $-4 \leq x \leq -2$ and $2 \leq y \leq 4$ so we can say that $\frac{y}{x} \leq 0$

can you finish the problem……..

Therefore, the expression $1+\frac{y}x$ will be maximized when $\frac{y}{x}$ is minimized, which occurs when $|x|$ is the largest and $|y|$ is the smallest.

can you finish the problem……..

Therefore in the region $(-4,2)$ , $\frac{x+y}{x}=1-\frac{1}{2}=\frac{1}{2}$