The formula for binomial theorem is
\((a+b)^n=( \left \ {{n} \atop {0}} \right.)a^{n}+( \left \ {{n} \atop {1}} \right.)a^{n-1}b+( \left \ {{n} \atop {0}} \right.)a^{n-2}b^2+...+( \left \ {{n} \atop {n}} \right.)b^n\)
Now this shall be very easy if the value of a = 1
The formula shall become
\((1+b)^n=( \left \ {{n} \atop {0}} \right.)1^{n}+( \left \ {{n} \atop {1}} \right.)1^{n-1}b+( \left \ {{n} \atop {0}} \right.)1^{n-2}b^2+...+( \left \ {{n} \atop {n}} \right.)b^n\)
Which shall be
\((1+b)^n=( \left \ {{n} \atop {0}} \right.)+( \left \ {{n} \atop {1}} \right.)b+( \left \ {{n} \atop {0}} \right.)b^2+...+( \left \ {{n} \atop {n}} \right.)b^n\)
So to find 11^4
We must break it as 1 + 10.
Option B) is the right answer.
