Question

If ${}^{100}{C}_{50}$ can be prime factorised as $2^{\alpha }\cdot 3^{\beta }\cdot 5^{\gamma }\cdot 7^{\delta }\dots$ where $\alpha ,\beta ,\gamma ,\delta ,\dots$ are non-negative integers, then the CORRECT relation(s) is (are)

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

Answer 1

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

$\gamma +\delta =0$

${}^{100}{C}_{50}=\frac{\left(100\right)!}{(50!)(50!)}=2^{\alpha }\cdot 3^{\beta }\cdot 5^{\gamma }\cdot 7^{\delta }\dots$ where $\alpha ,\beta ,\gamma ,\delta ,\dots$ are non- negative integers.
Exponent of $2$ in $\left(100\right)!$ is
$\left[\frac{100}{2}\right]+\left[\frac{100}{2^2}\right]+\left[\frac{100}{2^3}\right]+\cdots +\left[\frac{100}{2^6}\right]$

$=50+25+12+6+3+1=97$

Exponent of $2$ in $\left(50\right)!$ is
$\left[\frac{50}{2}\right]+\left[\frac{50}{2^2}\right]+\left[\frac{50}{2^3}\right]+\cdots +\left[\frac{50}{2^5}\right]$

$=25+12+6+3+1=47$
Hence, exponent of $2$ in ${}^{100}{C}_{50}$ is $3.$
$\{\because {}^{100}{C}_{50}=\frac{2^{97}}{2^{47}\cdot 2^{47}}I=2^{3}I\}$

Exponent of $3$ in $\left(100\right)!$
$\left[\frac{100}{3}\right]+\left[\frac{100}{3^2}\right]+\left[\frac{100}{3^3}\right]+\left[\frac{100}{3^4}\right]$

$=33+11+3+1=48$

Exponent of $3$ in $\left(50\right)!$ is
$\left[\frac{50}{3}\right]+\left[\frac{50}{3^2}\right]+\left[\frac{50}{3^3}\right]+\left[\frac{50}{3^4}\right]$

$=16+5+1=22$
Hence, exponent of $3$ in ${}^{100}{C}_{50}$ is $4.$

Exponent of $5$ in $\left(100\right)!$ is
$\left[\frac{100}{5}\right]+\left[\frac{100}{5^2}\right]+\left[\frac{100}{5^3}\right]$

$=20+4+0=24$

Exponent of $5$ in $\left(50\right)!$ is
$\left[\frac{50}{5}\right]+\left[\frac{50}{5^2}\right]+\left[\frac{50}{5^3}\right]$

$=10+2+0=12$
Hence, exponent of $5$ in ${}^{100}{C}_{50}$ is $0.$

Exponent of $7$ in $\left(100\right)!$ is
$\left[\frac{100}{7}\right]+\left[\frac{100}{7^2}\right]+\left[\frac{100}{7^3}\right]$

$=14+2+0=16$

Exponent of $7$ in $\left(50\right)!$ is
$\left[\frac{50}{7}\right]+\left[\frac{50}{7^2}\right]+\left[\frac{50}{7^3}\right]$

$=7+1+0=8$
Hence, exponent of $7$ in ${}^{100}{C}_{50}$ is $0.$

Exponent of $2,3,5,7$ in ${}^{100}{C}_{50}$ are $3,4,0,0$ respectively.
$\Rightarrow \alpha =3,\beta =4,\gamma =0,\delta =0$