Question

The sum $\sum _{0\le i}^{}\sum _{j\le 10}^{}\left({}^{10}{C}_j\right)\left({}^j{C}_i\right)$ is equal to

• A. $2^{10}-1$
• B. $2^{10}$
• C. $3^{10}-1$
• D. $3^{10}$

Answer 1

Answer: (D)

$3^{10}$

$\sum _{0\le i}^{}\sum _{j\le 10}^{}\left({}^{10}{C}_j\right)\left({}^j{C}_i\right)$

${=}^{10}{C}_0{(}^0{C}_0\left){+}^{10}{C}_1{(}^1{C}_1{+}^1{C}_0\right)+\dots \dots .{+}^{10}{C}_{10}{(}^{10}{C}_{10}{+}^{10}{C}_9+...\dots {+}^{10}{C}_0)$

$=2^0{.}^{10}{C}_0+2^1{.}^{10}{C}_1+\dots \dots ..+2^{10}{.}^{10}{C}_{10}$

Now,

$(1+2x{)}^{10}{=}^{10}{C}_0(1{)}^{10}{+}^{10}{C}_1\left(1{)}^9\right(2x{)}^1+\dots \dots \dots \dots .{+}^{10}{C}_{10}(2x{)}^{10}$

Put x = 1,

$\left(3{)}^{10}{=}^{10}{C}_0\right(1{)}^{10}{+}^{10}{C}_1\left(1{)}^9\right(2{)}^1+\dots \dots \dots \dots .{+}^{10}{C}_{10}(2{)}^{10}$

Thus the required sum is $3^{10}$