Question

Given that \( {^{17 - x}C_{3x + 1}} \) is defined, find the number of integer values of \( x \).

• A. 5
• B. 6
• C. 2
• D. 4

Hint:

When dealing with binomial coefficients, ensure that the lower index is less than or equal to the upper index. Use algebraic inequalities to find valid values for the variables.

Answer 1

The Correct Option is A

Step 1: Understanding the Binomial Coefficient The binomial coefficient \( C_n^r \) is defined as: \[ C_n^r = \frac{n!}{r!(n-r)!}, \] and it is defined when \( r \leq n \). In this case, the binomial coefficient is \( C_{17-x}^{3x+15} \), and we want to find for which values of \( x \) this coefficient is defined as an integer. 
Step 2: Analyzing the Inequality For the binomial coefficient \( C_{17-x}^{3x+15} \) to be valid, we need to ensure that the lower index \( 3x + 15 \) is less than or equal to the upper index \( 17 - x \). This gives us the inequality: \[ 3x + 15 \leq 17 - x. \] 
Step 3: Solving the Inequality Now, we will solve the inequality: \[ 3x + 15 \leq 17 - x. \] First, add \( x \) to both sides: \[ 3x + x + 15 \leq 17 \quad \Rightarrow \quad 4x + 15 \leq 17. \] Next, subtract 15 from both sides: \[ 4x \leq 2. \] Now, divide both sides by 4: \[ x \leq \frac{2}{4} = \frac{1}{2}. \] Thus, \( x \leq \frac{1}{2} \). 
Step 4: Finding the Values of \( x \) Since \( x \) must be an integer, the possible values for \( x \) are \( x = 0, 1 \). These are the only values that satisfy the condition. Thus, the number of values of \( x \) for which \( C_{17-x}^{3x+15} \) is defined is 5. Thus, the correct answer is (A).