Question

The greatest common divisor of ${}^{20}C_1,{}^{20}C_3,...............{}^{20}C_9$ is

• A. $20$
• B. $4$
• C. $5$
• D. None of these

Answer 1

The correct option is B

$4$

Hint: ${}^{n}C_r=\frac{n!}{(n-r)!r!}$,

Step 1: Find the value of ${}^{20}C_1$

$$\begin{gathered} {}^{20}C_1=\frac{20!}{19!1!} \\ \Rightarrow {}^{20}C_1=20 \end{gathered}$$

$\Rightarrow$ $20=2\times 2\times 5$

Step 2: Find the value of ${}^{20}C_3$

${}^{20}C_3=\frac{20!}{17!\times 3!}$

$\Rightarrow {}^{20}C_3=1140$

$1140=2\times 2\times 3\times 5\times 19$

Step 3: Find the value of ${}^{20}C_5$

${}^{20}C_5=\frac{20!}{15!\times 5!}$

$\Rightarrow {}^{20}C_5=15504$

$\Rightarrow$$15504=2\times 2\times 2\times 2\times 3\times 17\times 19$

Step 4: Find the value of ${}^{20}C_7$

$$\begin{gathered} {}^{20}C_7=\frac{20!}{13!\times 7!} \\ \Rightarrow {}^{20}C_7=77520 \end{gathered}$$

$\Rightarrow$$77520=2\times 2\times 2\times 2\times 3\times 5\times 17\times 19$

Step 5: Find the value of ${}^{20}C_9$

$$\begin{gathered} {}^{20}C_9=\frac{20!}{11!\times 9!} \\ \Rightarrow {}^{20}C_7=167960 \\ \Rightarrow 167960=2\times 2\times 2\times 5\times 13\times 17\times 19 \end{gathered}$$

From the prime factors of the numbers greatest common factor is $\left(2\times 2\right)$,

Therefore, greatest common divisor is $4$.

Hence, option $B$ is the correct answer.