Club A: The Home Favorites

Club A has been a dominant force in their domestic league and brings that momentum into the Champions League. With a strong squad depth and tactical flexibility, they are favorites in most of their home fixtures.

Key Players to Watch

• Maria Silva - Striker known for her clinical finishing and leadership on the field.
• Jin Soo Park - Midfielder with exceptional vision and playmaking abilities.

Strengths and Weaknesses

• Strengths: Strong defense, effective counter-attacks.
• Weakenesses: Vulnerable to high pressing teams.

Club B: The Consistent Contenders

Aiming to replicate their domestic success on the international stage, Club B has shown consistency in their performances against top-tier teams.

Key Players to Watch

• Lena Schmidt - Versatile defender capable of playing multiple positions effectively.
• Hana Kim - Goalkeeper with outstanding reflexes and shot-stopping ability.

Strengths and Weaknesses

• Strengths: Balanced squad, disciplined tactics.
• Weakenesses: Struggles against physically dominant teams.

Club C: The Dark Horse

Sometimes underestimated due to their recent league performance, Club C possesses young talent that could surprise many this season. Their resilience makes them a potential dark horse in Group E.

Key Players to Watch b+1 = a). - ( a = c+2 ) implies ( c = a-2), so ( b+1 > c+2 = a). - ( b+1 = c+2 ) implies ( b = c+1), so ( a > b+1 = c+2). 6. **Subtract Invalid Cases:** Calculate invalid cases where two variables are equal: - **Case 1:** ( a = b+1 ) implies ( b = a-1). Then: [ a + (a-1) + c = n-3 implies 2a + c = n-2 ] Solutions: ( (a, c) ) such that ( 2a + c = n-2 ). For each valid ( a ), ( c = n-2-2a) must be non-negative. - Range for ( a ): ( 0 leq a leq leftlfloor frac{n-2}{2} rightrfloor ). - Number of solutions: ( leftlfloor frac{n-2}{2} rightrfloor + 1 ). - **Case 2:** ( a = c+2 ) implies ( c = a-2). Then: [ a + b + (a-2) = n-3 implies 2a + b = n-1 ] Solutions: ( (a, b) ) such that ( 2a + b = n-1). For each valid ( a), ( b = n-1-2a) must be non-negative. - Range for ( a): ( 0 leq a leq leftlfloor frac{n-1}{2} rightrfloor ). - Number of solutions: ( leftlfloor frac{n-1}{2} rightrfloor + 1 ). - **Case 3:** ( b+1 = c+2) implies ( b = c+1). Then: [ a + (c+1) + c = n-3 implies a + 2c = n-4 ] Solutions: ( (a, c) ) such that ( a + 2c = n-4). For each valid ( c), ( a = n-4-2c) must be non-negative. - Range for ( c): ( 0 leq c leq leftlfloor frac{n-4}{2} rightrfloor ). - Number of solutions: ( leftlfloor frac{n-4}{2} rightrfloor + 1 ). 7. **Apply Inclusion-Exclusion Principle:** Subtract invalid cases from total: - Total solutions without distinctness constraint: ( binom{n-1}{2} ). - Subtract cases where two are equal: - Case 1: Subtract ( leftlfloor frac{n-2}{2} rightrfloor + 1). - Case 2: Subtract ( leftlfloor frac{n-1}{2} rightrfloor + 1). - Case 3: Subtract ( leftlfloor frac{n-4}{2} rightrfloor + 1). 8. **Final Formula:** The number of ways to choose distinct integers is: [ f(n) = binom{n-1}{2} - (leftlfloor frac{n-2}{2} rightrfloor + 1) - (leftlfloor frac{n-1}{2} rightrfloor + 1) - (leftlfloor frac{n-4}{2} rightrfloor + 1) This formula gives the number of ways to choose three distinct integers whose sum is exactly equal to any given integer value greater than or equal to six.