Question

Let $N=2^{200}\left({}^{200}{C}_0\right)\left({}^{200}{C}_{100}\right)-2^{198}\left({}^{200}{C}_1\right)\left({}^{199}{C}_{99}\right)+2^{196}\left({}^{200}{C}_2\right)\left({}^{198}{C}_{98}\right)+\cdots +\left({}^{200}{C}_{100}\right)\left({}^{100}{C}_0\right)$
then -

• A. $N$ is equal to the coefficient of $x^{100}$ in $(1-x{)}^{200}$
• B. The number of zeros at the end of $N$ is only one.
• C. $N=3^{101}\times k,$ where $k$ is an integer
• D. $N$ is equal to the number of ways of distributing $200$ different objects among $2$ persons equally.

Answer 1

Answer: (B), (C)

The correct options are
B The number of zeros at the end of $N$ is only one.
C $N=3^{101}\times k,$ where $k$ is an integer

$N=\sum _{r=0}^{100}(-1{)}^r.4^{100-r}{\cdot }^{200}{C}_r{\cdot }^{200-r}{C}_{100-r}$
$=\sum _{r=0}^{100}(-1{)}^r.4^{100-r}\times \frac{200!}{r!(200-r)!}\times \frac{(200-r)!}{(100-r)!100!}\times \frac{100!}{100!}$
$N={}^{200}{C}_{100}.(4-1{)}^{100}$
$N={}^{200}{C}_{100}.3^{100}$
no. of zero's in ${}^{200}{C}_{100}$ is equal to exponent of 5.
$\frac{200!}{100!100!}=\frac{(40+8+1)}{(2\times (20+4+0\left)\right)}$ = exp. of $5$ $\to$ $49-48=1$

Let's find the exponent of $3$ in ${}^{200}{C}_{100}$
${}^{200}{C}_{100}=\frac{200!}{100!\times 100!}{\text{exponent of 3 in}}^{200}{C}_{100}=\frac{\text{exponent of 3 in}200!}{\text{exponent of 3 in}100!\times 100!}=97-96=1\Rightarrow {}^{200}{C}_{100}{3}^{100}=k\times 3^{101}$

the number of ways of distributing $200$ different objects among $2$ persons equally $=\frac{200!}{100!100!}\times \frac{1}{2!}\times 2!={}^{200}{C}_{100}$

the coefficient of $x^{100}$ in $(1-x{)}^{200}={}^{200}{C}_{100}$