Question

If ${}^{100}{C}_{50}$ can be prime factorised as $2^{\alpha }\cdot 3^{\beta }\cdot 5^{\gamma }\cdot 7^b$ where $\alpha ,\beta ,\gamma ,\delta \dots \dots \dots \dots$ 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
${}^{100}{C}_{50}=\frac{(100)!}{(50)!(50)!}=2^{\alpha }\cdot 3^{\beta }\cdot 5^{\gamma }\cdot 7^{\delta }\dots \dots \dots$ where $\alpha ,\beta ,\gamma ,\delta \dots \dots$. 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.
$\therefore {}^{100}{C}_{50}=2^3\cdot 3^4\cdot 5^0\cdot 7^0\dots \dots \dots$
$\Rightarrow \alpha =3,\beta =4,\gamma =0,\delta =0$