If $A=\left\{\left(\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right): a_i, b_i, c_i,(i=1,2,3)\right.$ are the $\left.\begin{array}{l}\text { binomial coefficients in the expansion of }(1+\mathrm{x})^{11} \\ \text { then the number of elements in set } \mathrm{A} \text { is }\end{array}\right\}$,
$9^9$
$6^9$
$11^9$
$12^9$
Solution
$\because\left(1+\mathrm{x}^{11}\right)={ }^{11} \mathrm{C}_0 \mathrm{x}^0+{ }^{11} \mathrm{C}_1 \mathrm{x}^1+{ }^{11} \mathrm{C}_2 \mathrm{x}^2+{ }^{11} \mathrm{C}_3 \mathrm{x}^3$
$+{ }^{11} \mathrm{C}_4 \mathrm{x}^4+{ }^{11} \mathrm{C}_5 \mathrm{x}^5+{ }^{11} \mathrm{C}_6 \times \mathrm{x}^6+{ }^{11} \mathrm{C}_7 \mathrm{x}^7+{ }^{11} \mathrm{C}_8 \mathrm{x}^8+$
${ }^{11} \mathrm{C}_9 \mathrm{x}^9+{ }^{11} \mathrm{C}_{10} \mathrm{x}^{10}+{ }^{11} \mathrm{C}_{11} \mathrm{x}^{11}$
$\because{ }^{11} \mathrm{C}_0={ }^{11} \mathrm{C}_{11},{ }^{11} \mathrm{C}_1={ }^{11} \mathrm{C}_{10},{ }^{11} \mathrm{C}_2={ }^{11} \mathrm{C}_9,{ }^{11} \mathrm{C}_3$
$={ }^{11} \mathrm{C}_8,{ }^{11} \mathrm{C}_4={ }^{11} \mathrm{C}_7,{ }^{11} \mathrm{C}_5={ }^{11} \mathrm{C}_6$
So, In the expansion of $\left(1+x^{Y 1}\right)$, there are 6 different coefficients.
$\therefore \mathrm{a}_{\mathrm{i}}, \mathrm{b}_{\mathrm{i}}, \mathrm{c}_{\mathrm{i}}(\mathrm{i}=1,2,3)$ has 6 choices.
$\therefore$ No. of elements in the set $A=6^9$,