Question

${{}^{404}C}_4-{{}^{4}C}_1{{}^{303}C}_4+{{}^{4}C}_2{{}^{202}C}_4-{{}^{4}C}_3{{}^{101}C}_4$ is equal to

• A. ${\left(401\right)}^4$
• B. ${\left(101\right)}^4$
• C. $0$
• D. ${\left(201\right)}^4$

Answer 1

Answer: (B)

${\left(101\right)}^4$

The given expression here is the coefficient of $x^4$ in $P\left(x\right)=\left(\right(1+x{)}^{101}-1{)}^4$

$\Rightarrow$ $P\left(x\right)=\left(\right(1+x{)}^{101}-1{)}^4$

You can see this by first expanding $(y-1{)}^4$ then putting $y=(1+x{)}^{101}$ and expanding it in terms of $x$

But $P\left(x\right)=\left(\right(1+101x+...)-1{)}^4=(101x+...{)}^4={101}^4{x}^4(1+...{)}^4$, where the dots represent higher powers of $x$.

Thus, the coefficient of $x^4$ in $P\left(x\right)$ is ${101}^4$