Question

If ${ }^{100} C _{50}$ can be prime factorised as $2^\alpha \cdot 3^\beta \cdot 5^\gamma \cdot 7^{ b }$ where $\alpha, \beta, \gamma, \delta \ldots \ldots \ldots \ldots$ are non-negative integers, then correct relation is/are

• A. $\alpha<\beta$
• B. $\gamma<\delta$
• C. $\alpha+\delta=\beta+\gamma-1$
• D. $\gamma+\delta=0$

Answer 1

Answer: (A), (C), (D)

We have
${ }^{100} C _{50}=\frac{(100) !}{(50) !(50) !}=2^\alpha \cdot 3^\beta \cdot 5^\gamma \cdot 7^\delta \ldots \ldots \ldots$ where $\alpha, \beta, \gamma, \delta \ldots \ldots$. are non-negative integers.
Exponent of 2 in (100)!
$! \text { is } =\left[\frac{100}{2}\right]+\left[\frac{100}{2^2}\right]+\left[\frac{100}{2^3}\right]+\left[\frac{100}{2^4}\right]+\left[\frac{100}{2^5}\right]+\left[\frac{100}{2^6}\right] $
$ =50+25+12+6+3+1=97$
Exponent of 2 in $(50)$ !
$\text { is } =\left[\frac{50}{2}\right]+\left[\frac{50}{2^2}\right]+\left[\frac{50}{2^3}\right]+\left[\frac{50}{2^4}\right]+\left[\frac{50}{2^5}\right] $
$ =25+12+6+3+1=47$
Exponent of 2 in ${ }^{100} C _{50}$ is $=3$
$\left\{\because{ }^{100} C_{50}=\frac{2^{97}}{2^{47} \cdot 2^{47}} I=2^{3 \cdot} I\right\}$
In the similar way exponent of 3,5 and 7 in ${ }^{100} C _{50}$ are 4,0 and 0 respectively.
$\therefore{ }^{100} C _{50}=2^3 \cdot 3^4 \cdot 5^0 \cdot 7^0 \ldots \ldots \ldots $
$\Rightarrow \alpha=3, \beta=4, \gamma=0, \delta=0$