Question

use the factorial formula for combinations to prove that in general, ( _nc_r = _nc_{n-r} ).

using this formula, what is the formula for ( _nc_{n-r} )?

a. ( _nc_{n-r} = \frac{n!}{(n - r)!(r - (n - r))!} )

b. ( _nc_{n-r} = \frac{(n - r)!}{(n - r)!(n - (n - r))!} )

c. ( _nc_{n-r} = \frac{n!}{(n - r)!(n - (n - r))!} )

d. ( _nc_{n-r} = \frac{(n - r)!}{n!(n - r)!} )

Explanation:

Step1: Recall combination formula

The formula for combinations \( _nC_r \) is \( \frac{n!}{r!(n - r)!} \).

Step2: Substitute \( r \) with \( n - r \)

For \( _nC_{n - r} \), we substitute \( r \) in the combination formula with \( n - r \). So we get \( _nC_{n - r}=\frac{n!}{(n - r)!(n-(n - r))!} \).

Step3: Simplify the denominator

Simplify \( n-(n - r) \), which is \( n - n + r=r \). But we can also just look at the form. Let's check each option:

• Option A: The second factorial in denominator is \( (r-(n - r))!=(2r - n)! \), which is not correct.
• Option B: The numerator is \( (n - r)! \), but combination formula has numerator \( n! \), so incorrect.
• Option C: Matches the formula we derived by substituting \( r \) with \( n - r \) in \( _nC_r \) formula.
• Option D: Numerator and denominator are incorrect as per combination formula.

Answer:

C. \( \boldsymbol{_nC_{n - r}=\frac{n!}{(n - r)!(n-(n - r))!}} \)