Question

The value of $(21C_1-10C_1)+({}^{21}{C}_2-10C_2)+(21C_3-10C_3)+(21C_4-10C_4)+\dots .+(21C_{10}-10c_{10})$ is ?

• A. $2^{20}-2^{10}$
• B. $2^{21}-2^{11}$
• C. $2^{21}-2^{10}$
• D. $2^{20}-2^9$

Answer 1

Answer: (A)

$2^{20}-2^{10}$

Given,

to find the value of

$({}^{21}{C}_1-{}^{10}{C}_1)+{(}^{21}{C}_2-{}^{10}{C}_2)+({}^{21}{C}_3{-}^{10}{C}_3)+{(}^{21}{C}_4-{}^{10}{C}_4)+...\dots .+{(}^{21}{C}_{10}-{}^{10}{C}_{10})$

it becomes,

${(}^{21}{C}_1+{}^{21}{C}_2+{}^{21}{C}_3+...\dots +{}^{21}{C}_{10})$ $-{(}^{10}{C}_1+{}^{10}{C}_2+{}^{10}{C}_3+...\dots ..+{}^{10}{C}_{10})$

Let us multiply with $1$ & $-1$ we get,

$(1-1)+[{(}^{21}{C}_1+{}^{21}{C}_2+{}^{21}{C}_3+...\dots .+{}^{21}{C}_{10})-{(}^{10}{C}_1+{}^{10}{C}_2+{}^{10}{C}_3+...\dots .+{}^{10}{C}_{10}\left)\right]$

$({\because }^n{C}_o=1)$

Here, we get

${(}^{21}{C}_0-{}^{10}{C}_0)$ $[{(}^{21}{C}_1+{}^{21}{C}_2+{}^{21}{C}_3+...\dots ..+{}^{21}{C}_{10})-{(}^{10}{C}_1+{}^{10}{C}_2+{}^{10}{C}_3+...\dots +{}^{10}{C}_{10}\left)\right]$

$\Rightarrow {(}^{21}{C}_0+{}^{21}{C}_1+{}^{21}{C}_2+{}^{21}{C}_3+...\dots .+{}^{21}{C}_{10})-{(}^{10}{C}_0+{}^{10}{C}_1+{}^{10}{C}_2+{}^{10}{C}_3+...\dots ..+{}^{10}{C}_{10})$

$\because$ We know that,

$(1+x{)}^n={}^n{C}_0+{}^n{C}_1.x+{}^n{C}_2.x^2+...\dots \dots +{}^n{C}_n{x}^n$

For $x=1$, we get,

$(1+1{)}^n={}^n{C}_0+{}^n{C}_1+{}^n{C}_2+...\dots \dots {}^n{C}_n$

$\Rightarrow 2^n={}^n{C}_0+{}^n{C}_1+{}^n{C}_2+...\dots .{}^n{C}_n$

Here from $0$ to $10$ we have $11$ terms and for $n=21$ we get $22$ terms it mean half of the $\left\{0to10\right\}$ terms.

So, here we get,

$2^{21}={}^{21}{C}_0+...\dots .+{}^{21}{C}_{21}$

$2^{11}={}^{21}{C}_0+...\dots +{}^{21}{C}_{11}$

$2^{10}={}^{10}{C}_0+...\dots .+{}^{10}{C}_{10}$

$\therefore \frac{2^{21}}{2}-2^{10}$

$\Rightarrow 2^{20}-2^{10}$

$\therefore$ The value of ${(}^{21}{C}_1-{}^{10}{C}_1)+{(}^{21}{C}_2-{}^{10}{C}_2)+{(}^{21}{C}_3-{}^{10}{C}_3)+{(}^{21}{C}_{10}-{}^{10}{C}_{10})$

$=2^{20}-2^{10}$.