CAT Previous Year Questions| Permutation & Combination

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.