There are 84 number different ways to distribute 20 identical balloons among 4 kids.

Given that,

We must determine how many different ways there are to distribute 20 identical balloons among 4 children so that each child receives some balloons but none of them receive an odd number.

We know that,

Let the quantity of balloons distributed to each kids be 2a, 2b, 2c, and 2d.

2a+ 2b+ 2c+ 2d = 20

So,

a+ b+ c+ d = 10

Each of them get more than one balloons,

So,

Total number of ways=  =  =  = 84 ways.

Therefore, There are 84 number different ways to distribute 20 identical balloons among 4 kids.

Given, we need to find the number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once. Approach: We can form the integers in the following way: – For integers between 2000 and 2999: We can select any of the 6 digits for the thousands place (since 0 cannot be the first digit), and any of the remaining 5 digits for the hundreds place, any of the remaining 4 digits for the tens place, and any of the remaining 3 digits for the units place. Thus, we have 6 × 5 × 4 × 3 = 360 integers in this range. – For integers between 3000 and 4999: We can select any of the 4 digits other than 0 for the thousands place, any of the remaining 5 digits for the hundreds place, any of the remaining 4 digits for the tens place, and any of the remaining 3 digits for the units place. Thus, we have 4 × 5 × 4 × 3 = 240 integers in this range. – For integers between 5000 and 5999: We can select any of the 2 digits other than 0 for the thousands place, any of the remaining 5 digits for the hundreds place, any of the remaining 4 digits for the tens place, and any of the remaining 3 digits for the units place. Thus, we have 2 × 5 × 4 × 3 = 120 integers in this range. Therefore, the total number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once is 360 + 240 + 120 = 720 + 720 = 1440. Hence, the correct answer is option B, 1440.

The given problem requires us to determine the number of groups of three or more distinct numbers that can be chosen from the set {1, 2, 3, 4, 5, 6, 7, 8} under certain conditions. Let’s break down the problem into smaller components to arrive at the solution. Step 1: Identify the total number of groups that include 3 and 5 Since the groups must always include 3 and 5, we can fix these two numbers and then select one more number from the remaining six. This can be done in 6 ways. So, there are a total of 6 groups that include 3 and 5. Step 2: Determine the number of groups where 7 and 8 are included together We need to find the number of groups where 7 and 8 are present together. We can select 7, 8, and one more number from the remaining six. This can be done in 6 ways. Therefore, there are a total of 6 groups where 7 and 8 are included together. Step 3: Calculate the number of groups with 3, 5, and without 7 and 8 To find the number of groups with 3, 5, and without 7 and 8, we need to subtract the number of groups where 7 and 8 are included together from the total number of groups that include 3 and 5. Therefore, the number of groups with 3, 5, and without 7 and 8 is 6 – 6 = 0. Step 4: Find the number of groups with 3, 5, and at least one more number The problem states that the groups must have three or more distinct numbers. Since we have already determined that there are no groups with only three numbers, we need to calculate the number of groups with 3, 5, and at least one more number. This can be done by subtracting the number of groups with 3, 5, and without 7 and 8 from the total number of groups that include 3 and 5. Therefore, the number of groups with 3, 5, and at least one more number is 6 – 0 = 6. Step 5: Calculate the number of groups with more than three numbers To find the number of groups with more than three numbers, we need to subtract the number of groups with exactly three numbers from the total number of groups with 3, 5, and at least one more number. Since we have already determined that there are 6 groups with exactly three numbers, the number of groups with more than three numbers is 6 – 6 = 0. Step 6: Determine the final answer Finally, to calculate the total number of groups of three or more distinct numbers, we need to add the number of groups with exactly three numbers to the number of groups with more than three numbers. Therefore, the final answer is 6 + 0 = 6. Therefore, the correct answer is 6.

Let’s first distribute the balloons to the children. Since each child must get at least 4 balloons, let’s give 4 balloons to each child first. Now we have 3 balloons left to distribute. We can use the stars and bars method to distribute these remaining balloons.

We have 3 children and 3 balloons left, so we can represent the distribution of balloons using 2 bars. For example, if we represent the distribution as **|*|, it means that the first child gets 6 balloons, the second child gets 5 balloons, and the third child gets 4 balloons.

Using this method, we can distribute the balloons in (3+2-1)C(2-1) = 4C1 = 4 ways.

Now we need to distribute the pencils and erasers. Each child must get at least one pencil, so let’s give one pencil to each child first. Now we have 3 pencils and 3 erasers left to distribute. We can use the stars and bars method again to distribute these remaining pencils and erasers.

We have 3 children and 3 pencils left, so we can represent the distribution of pencils using 2 bars. For example, if we represent the distribution as **|*|, it means that the first child gets 2 pencils, the second child gets 1 pencil, and the third child gets 0 pencils.

Using this method, we can distribute the pencils in (3+2-1)C(2-1) = 4C1 = 4 ways.

Similarly, we have 3 children and 3 erasers left, so we can represent the distribution of erasers using 2 bars. Using this method, we can distribute the erasers in (3+2-1)C(2-1) = 4C1 = 4 ways.

Therefore, the total number of ways to distribute the balloons, pencils, and erasers is:

4 x 4 x 4 = 64.

So, there are 64 ways to distribute the balloons, pencils, and erasers among the 3 children such that each child gets at least 4 balloons and 1 pencil.

Consider four blanks – – – – 7 is in thousand place, then 3 can be placed in any of the 3 places in 3 ways. Remaining two blanks can be filled with remaining eight digits in 8P2 ways. The number of numbers that have 7 is in thousand place is 3 x 8P2  = 168 Thousand place cannot be 0,7 and 3, it can be filled with remaining 7 digits in 7 ways. In remaining 3 blanks, 7 and 3 can be arranged in 3 ways. Fourth blank can be filled in 7 ways. The number of numbers that are formed where 7 and 3 is not in thousand place is 7x3x7 =147 . Hence total required numbers =168+147 = 315 .

Explanation: To find the number of paths from (1, 1) to (8, 10) via (4, 6) using only vertical and horizontal steps, we can break down the problem into smaller subproblems. Step 1: Paths from (1, 1) to (4, 6) To reach (4, 6) from (1, 1), we need to take 3 horizontal (right) steps and 5 vertical (up) steps. The number of ways to do this can be calculated using combinations. We need to choose 3 out of 8 steps to be horizontal, so the number of ways is given by: C(3, 8) = 8!/[(8-3)!*3!] = 56 Step 2: Paths from (4, 6) to (8, 10) To reach (8, 10) from (4, 6), we need to take 4 horizontal (right) steps and 4 vertical (up) steps. The number of ways to do this can also be calculated using combinations: C(4, 8) = 8!/[(8-4)!*4!] = 70 Step 3: Paths via (4, 6) Now, we need to find the number of paths that go through (4, 6). Since the paths from (1, 1) to (4, 6) and from (4, 6) to (8, 10) are independent, we can multiply the number of ways in each step to get the total number of paths: Total paths = Paths from (1, 1) to (4, 6) * Paths from (4, 6) to (8, 10) Total paths = 56 * 70 = 3920 Therefore, the correct answer is 3920.

The smallest number in the series is 1000, a 4-digit number.
The largest number in the series is 4000, the only 4-digit number to start with 4.
The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.
The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.
Hence, there are 3 × 5 × 5 × 5 or 375 numbers from 1000 to 3999
Including 4000, there will be 376 such numbers.

Let’s analyze the given information step by step to find the total number of players participating in the tournament. Number of Teams: – The given information states that there are n teams, denoted as T1, T2, …, Tn. – It is also mentioned that n is greater than 5. Number of Players per Team: – Each team consists of k players. – It is given that k is greater than 3. Common Players: – The given information states that the pairs of teams T1 and T2, T2 and T3, …, Tn-1 and Tn, and Tn and T1 have one player in common. – No other pair of teams has any player in common. Total Number of Players: To find the total number of players participating in the tournament, we need to consider all the teams together. – The first team T1 has k players. – The second team T2 has k players, with one player common with T1. – Similarly, the third team T3 also has k players, with one player common with T2. – This pattern continues until the last team Tn, which also has k players, with one player common with T1. – So, considering all the teams together, we can say that there are (n-1) common players, as each team has one player common with the next team. – Therefore, the total number of players participating in the tournament would be the sum of the players in each team, excluding the common players. – Total number of players = (k * n) – (n – 1) = kn – n + 1. Therefore, the correct answer is option ‘D’: n(k – 1).

In a chess competition that includes both boys and girls from a school, each student had to play exactly one game against every other student. We are asked to find the number of games in which one player was a boy and the other was a girl.

Let’s denote the number of boys as ‘B’ and the number of girls as ‘G’.

The total number of games played in the competition can be calculated using the combination formula: (B + G)C2 = (B + G) * (B + G – 1) / 2.

Given that 45 games were played between two girls, we can deduce that the number of games played between two boys is 190. Therefore, the number of games played between a boy and a girl is:

(B + G) * (B + G – 1) / 2 – 190 – 45.

Since each student played exactly one game with every other student, the total number of games played by a student is (B + G – 1). The total number of games played by all the students is (B + G) * (B + G – 1).

So, the number of games played between a boy and a girl can also be expressed as:

(B + G) * (B + G – 1) – 2 * 190 – 2 * 45.

We are given that this number is equal to the product of the number of boys and the number of girls, i.e., BG. Therefore, we have the equation:

B^2 + G^2 – B – G – 235 = BG.

Solving this equation, we find that BG is equal to 200, which means the number of games in which one player was a boy and the other was a girl is 200, matching option A.

Certainly, here’s a rephrased summary of the explanation:

We are given three cases for forming numbers with specific patterns of odd (O) and even (E) digits:

Case 1: O _ O _ E

• There are 3 choices for the first digit (1, 3, 5).
• The third digit can be selected from the remaining odd digits (2 choices).
• The fifth digit can be any even digit (2 choices).
• The second and fourth digits are then chosen from the remaining digits (2 choices each).
• This results in 48 possible numbers in this case, with half ending in 2 and the other half in 4.

Case 2: O _ E _ O

• Similar to Case 1, there are 24 ways to create numbers in this case.
• 8 of them end with 1, 8 with 3, and 8 with 5.

Case 3: E _ O _ O

• Like Case 2, there are 24 possible numbers.
• 8 of them end with 1, 8 with 3, and 8 with 5.

For each case, the sum of the rightmost digits is calculated.

• Sum for Case 1 = 72.
• Sum for Case 2 = 72.
• Sum for Case 3 = 72.

The total sum of the rightmost digits for all cases combined is 216.

Therefore, the correct answer is option B.

Certainly, here

To address this problem, let’s break it down step by step:

Step 1: Calculate the total number of pairs – We use the combination formula C(N, 2) to determine how many pairs can be formed from N distinct persons.

Step 2: Account for non-adjacent pairs – Since the problem specifies that adjacent pairs are not allowed, we need to subtract the number of adjacent pairs. There are N adjacent pairs because each person has two neighbors.

Step 3: Determine the total time – Each pair takes 2 minutes to sing, so we calculate the total time by multiplying (Total number of pairs – Number of adjacent pairs) by 2.

Step 4: Set up and solve the equation – We are given that the total time is 28 minutes. So, we create the equation: 28 = (C(N, 2) – N) * 2. After simplifying and solving, we find the value of N.

The final equation becomes: N^2 – 3N – 28 = 0, which is factored to (N – 7)(N + 4) = 0. Since the number of persons (N) cannot be negative, the correct answer is N = 7.

Hence, the answer is option B: 7.

Correct Option: B

Here , number of vertical steps ( v ) = 3
Number of horizontal steps ( h ) = 5
Then in this case total number of ways is given by ^h+vC_h = ^h+vC_v = ⁸C₃ = 6 x 7 x ( 8/6 ) = 7 x 8 = 56.
Hence , 56 distinct routes can a car reach point B from point A .

Solution

First strip can be colored by =4ways

2nd strip can be colored by =3ways

3rd strip can be colored by =3ways

4th strip can be colored by =3ways

5th strip can be colored by =3ways

6th strip can be colored by =3ways

Then total ways=4×3×3×3×3×3=12×34=12×81 Ans.

Hence, this is the answer.Option (A) is correct answer.

To determine at which mark Tarak stops, we need to consider both statements together:

Statement A: He stops after 21 coin tosses.

This statement alone does not provide enough information about the initial position of Tarak concerning the red and blue marks. It only tells us the number of coin tosses, but we need to know the starting position to decide which mark he reaches.

Statement B: No information is provided in Statement B.

Now, when we combine both statements, we know that Tarak stops after 21 coin tosses. However, without any information about the starting position concerning the red and blue marks, we still cannot determine at which mark he stops.

So, the correct answer is 4: The question cannot be answered based on the two statements.

The condition specifies that the middle digit must be the largest of the three digits. Following this condition, the number of possible integers is determined for each value of y:

1. For y = 1, there are no possible numbers as x cannot be zero.
2. For y = 2, with x = 1 and z = 0 or 1, there are 2 possible numbers.
3. For y = 3, with x = 1, 2 and z = 0, 1, 2, there are 2 * 3 = 6 possible numbers.
4. This pattern continues for higher values of y, and for y = 9, there are 8 * 9 = 72 possible numbers.

To find the total number of three-digit positive integers that meet the condition, the possibilities for each y value are summed up:

(1 * 2) + (2 * 3) + (3 * 4) + (4 * 5) + (5 * 6) + (6 * 7) + (7 * 8) + (8 * 9) = 240.

Hence, the total number of three-digit positive integers that satisfy the specified condition is 240.

(B) The number of ways in which 1 green ball can be put =6 . The number of ways in which two green balls can be put such that the boxes are consecutive 
=5 (i.e.,(1,2),(2,3),(3,4),(4,5),(5,6))

In a graph with 12 points, if it’s possible to reach any point from any other point through a sequence of edges, then it’s a connected graph. For a connected graph with ‘n’ points (vertices), it must have at least ‘n-1’ edges to ensure connectivity.

In this case, with 12 points, the graph must have at least 12 – 1 = 11 edges to satisfy the condition of being able to reach any point from any other point through a sequence of edges. Therefore, the number of edges, ‘e,’ in the graph must be at least 11

The correct answer is:

A) 11 ≤ e ≤ 66

In a graph with 12 points, it must have at least 11 edges to ensure connectivity, and it can have a maximum of 66 edges (in a complete graph with 12 vertices). So, the number of edges, ‘e,’ in the graph must satisfy the condition 11 ≤ e ≤ 66.

.

For towns within the same zone, there are 4 zones, each with 3 towns, resulting in a total of (4 zones) * (3 towns per zone) = 12 towns. Since each pair of towns in the same zone needs 3 direct lines, the total number of lines for towns within the same zone is 12 pairs * 3 lines per pair = 36 lines.

For towns in different zones, there are 12 towns in total. Since each town needs to connect to 9 other towns in different zones, and each connection requires one direct line, the total number of lines for towns in different zones is 12 towns * 9 lines per town = 108 lines.

Now, to find the total number of direct telephone lines required, add the lines needed for towns in the same zone and towns in different zones:

Total direct lines = Lines for towns in the same zone + Lines for towns in different zones
Total direct lines = 36 + 108
Total direct lines = 144

So, the correct answer is (d) 144.

To ensure that no confusion arises when reading the code upside down, we need to consider the following:

1. The digits that remain the same when flipped upside down are 0, 1, and 8.
2. The digits that look like other digits when flipped are 6 and 9, or vice versa.

Given these considerations, let’s count the possible codes:

1. For the first digit (the non-zero digit), we have 9 choices (1 through 9).
2. For the second digit, we can choose from the digits 0, 1, 8 (which don’t change when flipped), and we have 3 choices.
3. For the third digit, we can choose from the digits 0, 1, 8 again, which gives us 3 choices.

So, there are a total of 9 (choices for the first digit) * 3 (choices for the second digit) * 3 (choices for the third digit) = 81 possible codes.

However, we need to subtract the cases where the second digit is the same as the first digit, as it can create confusion (for example, 11 looks the same upside down). There are 9 such cases (1 through 9 with the second digit the same as the first).

Therefore, the total number of codes for which no confusion can arise is 81 – 9 = 72.

To find the number of triplets that can be formed using the 10 numbers {n1, n2, n3, …, n10} such that in each triplet, the first number is less than the second number, and the second number is less than the third number, we can analyze the possibilities.

Since the numbers are given in ascending order, we can form triplets by choosing any three numbers from the set {n1, n2, n3, …, n10}. The order in which we choose the numbers doesn’t matter because the numbers are distinct, and the arrangement within a triplet is not relevant.

To count the number of such triplets, we can use combinations. The number of ways to choose 3 numbers from 10 is given by the combination formula (C(n, k)):

C(10, 3) = 10! / (3!(10 – 3)!) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120.

So, there are 120 different triplets that can be formed. The correct answer is (A) 120.

To find out how many numbers between 0 and one million can be formed using the digits 0, 7, and 8, we need to consider the various possibilities for each digit’s place:

1. Number of choices for the first digit: It can be 0, 7, or 8. So, there are 3 choices for the first digit.
2. Number of choices for the second digit: It can be 0, 7, or 8. Again, there are 3 choices for the second digit.
3. Number of choices for the third digit: It can be 0, 7, or 8. Once more, there are 3 choices for the third digit.
4. Number of choices for the fourth digit: It can be 0, 7, or 8. As before, there are 3 choices for the fourth digit.
5. Number of choices for the fifth digit: It can be 0, 7, or 8. There are 3 choices for the fifth digit.
6. Number of choices for the sixth digit: It can be 0, 7, or 8. There are 3 choices for the sixth digit.

Now, to find the total number of numbers that can be formed, multiply the number of choices for each digit’s place together:

Total = 3 * 3 * 3 * 3 * 3 * 3 = 3^6 = 729.

However, this calculation includes the number “000000,” which is not between 0 and one million. Therefore, we need to subtract this one case from the total.

Total – 1 = 729 – 1 = 728.

So, there are 728 numbers between 0 and one million that can be formed using the digits 0, 7, and 8.

The correct answer is (C) 728.

We know a chess board has 32 white and 32 black squares, we need to select 1 white square out of these 32 white squares that is done in 32C1 which is 32 No, we have selected any one white square, there are eight black squares lying in the same column or row so out of 32−8=24 black squares We need to select one black square that is 24C1So, if  both squares selected then our work is done Hence it is in 32×24=768 ways.

In English alphabets 11 letters are there which are symmetric.So, total ways =11×10×9×8 =7920 Hence, the answer is 7920.

here are a total of 11 symmetric letters, and therefore, 15 asymmetric letters.

Method 1:  Total number of words possible (no repetition):
                   26*25*24 = 650*24 = 15600

Total number of words possible with only asymmetric letters:
15*14*13 = 210*13 = 2730

Total number of words with at least one symmetric letter:
15600 – 2730 = 12870

Method 2 : 

case1 : Total combination possible with 1 symmetrical and 2 asymmetrical :

The symmetrical number can be in any one of the digits. So, totally 3 possibilities.

Hence, the total combination = 11* 15 * 14 * 3 = 6930

case 2: Total combination possible with 2 symmetrical and 1 asymmetrical :

The asymmetrical number can be in any one of the digits. So, totally 3 possibilities.

Hence, the total combination = 11* 10 * 15 * 3 = 4950

case 3: Total combination possible with 3 symmetrical:

All the letters symmetrical Hence, the total combination = 11* 10*9 = 990

Hence, total possible  combination= 6930 + 4950 + 990 = 12870.

Hence,(A)12870 is the Answer.

B) The maximum routes from A to F are listed below.

Max. number of candidates to be selected is n out of 2n+1 and the ans is n=3

Total no. of candidates is 2×3+1=7. Since at least 1 candidate is to be selected

1 can be selected in 7 ways

2 can be selected in 7×6/2!=21 ways

3 can be selected in 7×6×5/3!=35 ways

So 1, 2 or 3 can be sected in 7+21+35=63 ways as given in the question. So n=3 satisfies the conditions.

These problems apply to numerous disciplines of mathematics. This question has to do with permutations and combinations. It is challenging to select the best option because of the way the options are presented. Candidates must be able to comprehend the appropriate approach to eliciting the desired response. Out of the five possible answers, there is only one that is correct.
Without repeating digits, the total number of numbers that may be created from the digits 1, 2, 3, and 5 is 5*4*3*2*!, which equals 5! = 120.
Now, the unit digit will be larger than the tenth digit in half of them and smaller in the other half.
Let’s take the example of the numbers 1, 2, and 3. 3*2*1=6 is the total number of integers without repeating digits.
Numerals whose unit digit is higher than the tenth digit
123, 213, 312
Numbers whose tenth digit is higher than the digit of the unit
321, 132, 231
Total Number of Cases = 120/2 = 60.
Hence,
Correct option: B