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The number of solutions of x1 + x2 + x3 ...

The number of solutions of `x_1 + x_2 + x_3 = 51` (X1, X2, X3 being odd natural numbers) is

A

300

B

325

C

330

D

350

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the number of solutions for the equation \( x_1 + x_2 + x_3 = 51 \) where \( x_1, x_2, x_3 \) are odd natural numbers, we can follow these steps: ### Step 1: Express odd natural numbers in terms of integers Since \( x_1, x_2, x_3 \) are odd natural numbers, we can express them in the following form: \[ x_1 = 2a + 1, \quad x_2 = 2b + 1, \quad x_3 = 2c + 1 \] where \( a, b, c \) are non-negative integers. ### Step 2: Substitute into the equation Substituting these expressions into the equation gives: \[ (2a + 1) + (2b + 1) + (2c + 1) = 51 \] This simplifies to: \[ 2a + 2b + 2c + 3 = 51 \] ### Step 3: Simplify the equation Now, we can simplify this equation: \[ 2a + 2b + 2c = 51 - 3 \] \[ 2a + 2b + 2c = 48 \] Dividing the entire equation by 2, we get: \[ a + b + c = 24 \] ### Step 4: Find the number of non-negative integer solutions We need to find the number of non-negative integer solutions to the equation \( a + b + c = 24 \). This is a classic problem in combinatorics and can be solved using the "stars and bars" theorem. According to the stars and bars theorem, the number of solutions for the equation \( x_1 + x_2 + \ldots + x_r = n \) in non-negative integers is given by: \[ \binom{n + r - 1}{r - 1} \] In our case, \( n = 24 \) and \( r = 3 \) (since we have three variables: \( a, b, c \)). ### Step 5: Apply the formula Substituting the values into the formula gives: \[ \binom{24 + 3 - 1}{3 - 1} = \binom{26}{2} \] ### Step 6: Calculate \( \binom{26}{2} \) Now we calculate \( \binom{26}{2} \): \[ \binom{26}{2} = \frac{26 \times 25}{2 \times 1} = \frac{650}{2} = 325 \] ### Conclusion Thus, the number of solutions of \( x_1 + x_2 + x_3 = 51 \) where \( x_1, x_2, x_3 \) are odd natural numbers is: \[ \boxed{325} \]
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