# If α = Mc2, Then Find the Value of αC2. - Mathematics

If α = mC2, then find the value of αC2.

#### Solution

${}^\alpha C_2 = \frac{\alpha}{2} \times \frac{(\alpha - 1)}{1} \times^\alpha C_0$  [∵${}^n C_r = \frac{n}{r} . {}^{n - 1} C_{r - 1}$]

${}^n C_r = \frac{n}{r} . {}^{n - 1} C_{r - 1}$  [∵${}^n C_0 = 1$]
$= \frac{1}{2} \left[ {}^m C_2 \left( {}^m C_2 - 1 \right) \right]$
$= \frac{1}{2} \left[ \frac{m!}{2! \left( m - 2 \right)!} \left( \frac{m!}{2! \left( m - 2 \right)!} - 1 \right) \right]$
$= \frac{1}{2} \left[ \frac{m \left( m - 1 \right)}{2} \left( \frac{m \left( m - 1 \right)}{2} - 1 \right) \right]$
$= \frac{1}{2} \left[ \frac{m\left( m - 1 \right)}{2} \left( \frac{m \left( m - 1 \right) - 2}{2} \right) \right]$
$= \frac{1}{8} \left[ m \left( m - 1 \right) \left\{ m \left( m - 1 \right) - 2 \right\} \right]$
$= \frac{1}{8} \left[ m^2 \left( m - 1 \right)^2 - 2m \left( m - 1 \right) \right]$
$= \frac{1}{8} \left[ m^2 \left( m^2 + 1 - 2m \right) - 2 m^2 + 2m \right]$
$= \frac{1}{8} \left[ m^4 + m^2 - 2 m^3 - 2 m^2 + 2m \right]$
$= \frac{1}{8} \left[ m^4 - 2 m^3 - m^2 + 2m \right]$
$= \frac{1}{8} \left[ \left( m^2 - 2m \right) \left( m^2 - 1 \right) \right]$
$= \frac{1}{8} \left[ m \left( m - 2 \right) \left( m - 1 \right) \left( m + 1 \right) \right]$
$= \frac{1}{8} \left( m + 1 \right) m \left( m - 1 \right) \left( m - 2 \right)$
Concept: Combination
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 17 Combinations
Exercise 17.1 | Q 15 | Page 8