Permutations And Combinations

To solve the problem, we need to determine the domain of the function , where the binomial coefficient is defined only when the upper value is a non-negative integer and the lower value is a non-negative integer ≤ upper value.

1. Understand the Condition for Combinations:
The function involves a binomial coefficient , which is only defined when:
- (i.e., n is a non-negative integer)
- and

2. Apply to the Given Function:
We are given

3. Set Conditions for Domain:
For the binomial to be defined:
-
-
- Also,

4. Find Integer Values Satisfying All Conditions:
We need integer values of such that:
-
So, possible integer values are:

Final Answer:
The domain of the function is

The expression is defined and real for all where is non-negative.

Step 1: Simplify the expression and set up the inequality. First, simplify the expression: To ensure the base of the power is non-negative:

Step 2: Solve the inequality. To find the values of satisfying this inequality: Solving this inequality: Calculating the cube root:

Step 3: Refine the domain. Since cannot be zero, the domain of that satisfies the inequality, ensuring the entire expression is valid (not just non-negative), is:

The coefficient of in the polynomial expansion of is determined by identifying the term that matches the desired variables and their exponents. This is achieved through the multinomial expansion, where the specific term we are interested in is given by: Thus, the coefficient is 6480.

Step 1: The available digits are 1, 3, 5, 7, and 9. According to the divisibility rule for 3, a number is divisible by 3 if the sum of its digits is divisible by 3.

Step 2: The total number of 3-digit numbers that can be formed from the digits 1, 3, 5, 7, and 9 without repetition is: Now, we need to determine how many of these 3-digit numbers are divisible by 3.

Step 3: To check divisibility by 3, we need to ensure the sum of the digits is divisible by 3. Let's evaluate the possible sums of digits formed from the set {1, 3, 5, 7, 9}: 1 + 3 + 5 = 9 (divisible by 3)
1 + 3 + 7 = 11 (not divisible by 3)
1 + 3 + 9 = 13 (not divisible by 3)
1 + 5 + 7 = 13 (not divisible by 3)
1 + 5 + 9 = 15 (divisible by 3)
1 + 7 + 9 = 17 (not divisible by 3)
3 + 5 + 7 = 15 (divisible by 3)
3 + 5 + 9 = 17 (not divisible by 3)
3 + 7 + 9 = 19 (not divisible by 3)
5 + 7 + 9 = 21 (divisible by 3)
Step 4: The valid combinations of digits that result in a sum divisible by 3 are: {1, 3, 5}
{1, 5, 9}
{3, 5, 7}
{5, 7, 9}
Step 5: For each valid combination, the number of possible 3-digit numbers is the number of ways to arrange 3 distinct digits, which is .

Step 6: Since there are 4 valid combinations and each combination can be arranged in 6 ways, the total number of 3-digit numbers divisible by 3 is: Thus, the number of 3-digit numbers divisible by 3 is .

To calculate the sum of all 4-digit numbers formed from the digits 0, 3, 6, and 9 without repetition, we consider the contribution of each digit to the thousands, hundreds, tens, and units place.

Step 1: Determine valid permutations for each position. Since 0 cannot be the first digit of a 4-digit number, only the digits 3, 6, or 9 can occupy the first place. Each of the remaining three positions can be occupied by any of the remaining three digits.

Step 2: Calculate permutations and place value contributions. - Thousands place: Each of 3, 6, and 9 can appear in the thousands place in (6) different numbers:

- Hundreds, Tens, and Units places: Here, 0 can appear, contributing to (6) permutations per digit per place: Combining these, the total contribution for hundreds, tens, and units places together is:

Step 3: Sum all contributions. Add the contributions from all places: However, the correct approach considers all digits appearing times in each place value due to all permutations where they are non-zero. Correctly calculating based on digit positioning: Correcting for earlier miscalculation in the sum of permutations, we identify that the corrected sum is actually lower, adjusting based on all possible valid combinations of numbers: Then, subtracting the incorrect overestimate of zero's contributions (not contributing to the thousands place), we get:

Conclusion: The correct sum of all valid 4-digit numbers formed from 0, 3, 6, and 9 without repetition is 115992, matching option (2).

Step 1: The question paper consists of three sections: A, B, and C. We are tasked with determining the number of ways a candidate can answer exactly 7 questions, subject to the following conditions: - No more than 4 questions can be answered from part A. - No more than 3 questions can be answered from part B. - No more than 2 questions can be answered from part C. Let , , and represent the number of questions answered from parts A, B, and C, respectively. We know: Additionally, the constraints are:

Step 2: We now find the valid combinations for , , and that satisfy the given constraints. The possible combinations where the sum of the questions equals 7 are: - - - - - -

Step 3: For each valid combination, we calculate the number of ways to select the questions from each part: - For , the number of ways to choose the questions is: - For , the number of ways to choose the questions is: - For , the number of ways to choose the questions is: - For , the number of ways to choose the questions is: - For , the number of ways to choose the questions is: - For , the number of ways to choose the questions is:

Step 4: The total number of ways is obtained by summing the values calculated for each valid combination: Therefore, the total number of ways the candidate can answer exactly 7 questions is . \bigskip

Each of the 6 distinct items can be placed in one of the 2 boxes, giving a total of ways to distribute all items. However, this total includes the cases where one of the boxes is empty. We need to subtract these cases to ensure that both boxes contain at least one item. There are 2 scenarios where one box is empty: all items are in box 1 or all items are in box 2. Each scenario is counted once, so we subtract 2 from 64: Thus, there are 62 ways to distribute the 6 items into 2 boxes such that no box is empty.