Question

The coefficient of $x^4$ in the expansion of ${(1+x+x^2+x^3)}^6$ in powers of $x$, is

• A. $110$
• B. $120$
• C. $130$
• D. $100$

Answer 1

The correct option is B

$120$

Given${(1+x+x^2+x^3)}^6$

To find coefficient $x^4$

using permutation and combination to solve

Hence rearranging the given expression gives

$$\begin{gathered} {(1+x+x^2+x^3)}^n=(1+x{)}^n(1+x2{)}^n \\ {(1+x+x^2+x^3)}^6=(1+x{)}^6{(1+x^2)}^6 \end{gathered}$$

Coefficient of $x^4$

$$\begin{gathered} {\Rightarrow }^n{C_0}^n{C}_0{}^n{C}_2.{}^n{C}_1{+}^n{C}_4.{}^n{C}_6 \\ ={}^6{C}_0{}^6{C}_2+{}^6{C}_2.{}^6{C}_1{+}^6{C}_4.{}^6{C}_6 \\ =15+(15\times 6)+15 \\ =120 \end{gathered}$$

Hence option B is correct