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p, q and rare odd natural numbers such that p + q + r = 19 and p < q < r. The number of possible solution sets (p, q, r) is
    Correct answer is '5'. Can you explain this answer?
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    p, q and rare odd natural numbers such that p + q + r = 19 and p q r....
    (p, q, r) = (1, 3, 15), (1, 5, 13), (1, 7, 11), (3, 5, 11) and (3, 7, 9)
    There are 5 sets of (p, q, r) for which the conditions p < q < r and p + q + r = 19 are satisfied.
    Answer: 5
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    p, q and rare odd natural numbers such that p + q + r = 19 and p q r....
    Number of Possible Solution Sets for p, q, and r

    To find the number of possible solution sets for p, q, and r, we need to consider the given conditions: p + q + r = 19 and p < q="" />< />

    Let's analyze the problem step by step:

    1. Understanding the conditions:
    - We are given three natural numbers: p, q, and r.
    - The sum of these numbers is 19: p + q + r = 19.
    - The numbers are odd, i.e., they are not divisible by 2.
    - The numbers are distinct, i.e., they are not equal to each other.

    2. Analyzing the sum:
    Since the sum of the three numbers is odd (19), it implies that at least one of the numbers must be odd. Let's consider three cases based on the parity of the numbers:

    - Case 1: Only one number is odd.
    In this case, the sum of two even numbers would be even, which contradicts the fact that the sum of the three numbers is odd. Therefore, this case is not possible.

    - Case 2: Two numbers are odd.
    In this case, if the sum of the two odd numbers is even, then the third number (which is even) should be odd, which is not possible. Therefore, this case is also not possible.

    - Case 3: All three numbers are odd.
    This case satisfies all the conditions given in the problem statement.

    3. Analyzing the range of possible values:
    Since the numbers are distinct and odd, we can determine the possible range for each number:
    - p: The smallest odd natural number is 1, so p can take values from 1 to 17 (inclusive).
    - q: Since q should be greater than p, q can take values from p+2 to 19 (inclusive).
    - r: Since r should be greater than q, r can take values from q+2 to 19 (inclusive).

    4. Counting the solution sets:
    We need to count the number of possible combinations of p, q, and r that satisfy the given conditions. To do this, we can use nested loops to iterate over the possible values of p, q, and r.

    The number of possible solutions can be determined by counting the number of iterations in the nested loops. The loops can be implemented as follows:

    - Loop 1: Iterate over the possible values of p from 1 to 17.
    - Loop 2: Iterate over the possible values of q from p+2 to 19.
    - Loop 3: Iterate over the possible values of r from q+2 to 19.

    We can count the number of iterations for each loop and multiply them together to obtain the total number of possible solution sets.

    5. Calculating the number of iterations:
    - Loop 1: There are 17 possible values for p.
    - Loop 2: For each value of p, there are (19 - p - 2) possible values for q.
    - Loop 3: For each combination of p and q, there are (19 - q - 2) possible values for r.

    To calculate the total number of iterations, we multiply the number of iterations for each loop:
    Total iterations = 17 * (19 - p - 2) * (19 - q - 2)

    6. Evaluating the expression:
    We need to find the
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    p, q and rare odd natural numbers such that p + q + r = 19 and p q r. The number of possible solution sets (p, q, r) isCorrect answer is '5'. Can you explain this answer?
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