Question

What is the value of \( ^nC_{n-1} \)?

• A. \(1 \)
• B. \(n! \)
• C. \(n \)
• D. \(0 \)

Hint:

Remember the symmetry property of combinations: \( ^nC_r=\, ^nC_{n-r}\). Thus \( ^nC_{n-1}=\, ^nC_1=n\).

Answer 1

The Correct Option is C

Concept: In mathematics, combinations are used to determine the number of ways of selecting objects from a group when the order of selection does not matter. The general formula for combinations is: \[ ^nC_r=\frac{n!}{r!(n-r)!} \] where: \[ n!=n\times(n-1)\times(n-2)\times \cdots \times 1 \] One important property of combinations is symmetry, given by: \[ ^nC_r=\, ^nC_{n-r} \] This property helps simplify expressions involving combinations.

Step 1: Writing the standard formula for the given combination expression.
The expression given in the question is: \[ ^nC_{n-1} \] Using the combination formula: \[ ^nC_r=\frac{n!}{r!(n-r)!} \] Substituting: \[ ^nC_{n-1}=\frac{n!}{(n-1)!(n-(n-1))!} \]

Step 2: Simplifying the factorial expression carefully.
Now simplify the denominator. First: \[ n-(n-1)=1 \] So the expression becomes: \[ ^nC_{n-1}=\frac{n!}{(n-1)!1!} \] Since: \[ 1!=1 \] we obtain: \[ ^nC_{n-1}=\frac{n!}{(n-1)!} \]

Step 3: Expanding factorial terms to obtain the final result.
Expand the factorial: \[ n!=n\times(n-1)! \] Substituting this gives: \[ ^nC_{n-1}=\frac{n\times(n-1)!}{(n-1)!} \] Cancelling common terms: \[ ^nC_{n-1}=n \] Hence the required value becomes: \[ \boxed{n} \] Therefore option (C) is the correct answer.