Question 27

Mathematics Permutation and Combination Hard

If n and r are integers such that 1&nbsp;&le; r&nbsp;&le; n, then the value of&nbsp;<span class="math-tex">\(n(^{n-1}C_{r-1})\)</span>&nbsp;is

(A) <span class="math-tex">\(^nC_r\)</span>
(B) <span class="math-tex">\(r(^nC_r)\)</span>
(C) <span class="math-tex">\(n(^nC_r)\)</span>
(D) <span class="math-tex">\((n-1)(^nC_r)\)</span>
View Dynamic Solution & Explanation
Correct Solution: Option B

Step-by-step Solution:

To find the value of \( n \cdot {}^{(n-1)}C_{(r-1)} \), we can proceed with the following steps: 1. Express the Combination: The combination \( {}^{(n-1)}C_{(r-1)} \) is given by: \[ {}^{(n-1)}C_{(r-1)} = \frac{(n-1)!}{(r-1)! \cdot (n-r)!} \] 2. Multiply by \( n \): Multiply the combination by \( n \): \[ n \cdot {}^{(n-1)}C_{(r-1)} = n \cdot \frac{(n-1)!}{(r-1)! \cdot (n-r)!} \] 3. Simplify the Expression: Notice that \( n \cdot (n-1)! = n! \): \[ n \cdot {}^{(n-1)}C_{(r-1)} = \frac{n!}{(r-1)! \cdot (n-r)!} \] 4. Express in Terms of \( nC_r \): The combination \( nC_r \) is given by: \[ nC_r = \frac{n!}{r! \cdot (n-r)!} \] Notice that: \[ \frac{n!}{(r-1)! \cdot (n-r)!} = r \cdot \frac{n!}{r! \cdot (n-r)!} = r \cdot nC_r \] Therefore, the value of \( n \cdot {}^{(n-1)}C_{(r-1)} \) is: \[ \boxed{r \cdot nC_r} \]