Question

The coefficient of $$x^{7}$$ in $$\left( 1-x-x^{2}+x^{3}\right)^{6} $$ is
A. -144
B. 144
C. -128
D. -142

Answer 1

Hint: In this question it is given that we have to find the coefficient of $$x^{7}$$ from the expression $$\left( 1-x-x^{2}+x^{3}\right)^{6} $$.
So to find the solution we need to know about binomial expansion, which is $$\left( 1-a\right)^{n} =\ ^{n} C_{0}-\ ^{n} C_{1}\cdot a+\ ^{n} C_{2}\cdot a^{2}-\ldots +(-1)^{n}\ ^{n} C_{n}\cdot a^{n}$$.
So by this expression we are able to find the solution.

Complete step-by-step answer:
Here given expression,
$$\left( 1-x-x^{2}+x^{3}\right)^{6} $$
taking $x^2$ common from 3rd and 4th terms, we get.
$$=\left\{ \left( 1-x\right) -x^{2}\left( 1-x\right) \right\}^{6} $$
$$=\left\{ \left( 1-x\right) \left( 1-x^{2}\right) \right\}^{6} $$
$$=\left( 1-x\right)^{6} \left( 1-x^{2}\right)^{6} $$
By expanding the 1st and 2nd terms by the use of Binomial expansion, we get,
$$=\left( {}^{6}C_{0}-\ ^{6} C_{1}x+\ ^{6} C_{2}x^{2}-\ldots +\ ^{6} C_{6}x^{6}\right) \left( {}^{6}C_{0}-\ ^{6} C_{1}\left( {}x^{2}\right) +\ ^{6} C_{2}\left( x^{2}\right)^{2} -\ ^{6} C_{3}\left( x^{2}\right)^{3} +\ldots +\ ^{6} C_{6}\left( x^{2}\right)^{6} \right) $$
On simplifying the power of ‘x’, we get,
$$=\left( {}^{6}C_{0}-\ ^{6} C_{1}x+\ ^{6} C_{2}x^{2}-\ldots +\ ^{6} C_{6}x^{6}\right) \left( {}^{6}C_{0}-\ ^{6} C_{1}x^{2}+\ ^{6} C_{2}x^{4}-\ ^{6} C_{3}x^{6}+\ldots +\ ^{6} C_{6}x^{12}\right) $$
Now we have to find the coefficient of $$x^{7}$$,
Therefore, the coefficient of $$x^{7}$$ is
$$=\left( \text{coefficient of} \ x\times \text{coefficient of} \ x^{6}\right) +\left( \text{coefficient of} \ x^{3}\times \text{coefficient of} \ x^{4}\right) +\left( \text{coefficient of} \ x^{5}\times \text{coefficient of} \ x^{2}\right) $$
$$=\left\{ \left( -{}^{6}C_{1}\right) \times \left( -{}^{6}C_{3}\right) \right\} +\left\{ \left( -{}^{6}C_{3}\right) \times \left( {}^{6}C_{2}\right) \right\} +\left\{ \left( -{}^{6}C_{5}\right) \times \left(-{}^{6}C_{1}\right) \right\} $$
$$=\left( {}^{6}C_{1}\ ^{6} C_{3}\right) -\left( {}^{6}C_{3}\ ^{6} C_{2}\right) +\left( {}^{6}C_{5}\ ^{6} C_{1}\right) $$
We know that $${}^{n}C_{r}$$ can be written as, $${}^{n}C_{r}=\dfrac{n!}{r!\cdot \left( n-r\right) !}$$
$$\dfrac{6!}{1!\left( 6-1\right) !} \times \dfrac{6!}{3!\left( 6-3\right) !} -\dfrac{6!}{3!\left( 6-3\right) !} \times \dfrac{6!}{2!\left( 6-2\right) !} +\dfrac{6!}{5!\left( 6-5\right) !} \times \dfrac{6!}{1!\left( 6-1\right) !}$$
$$=\dfrac{6!}{1!\cdot 5!} \times \dfrac{6!}{3!\cdot 3!} -\dfrac{6!}{3!\cdot 3!} \times \dfrac{6!}{2!\cdot 4!} +\dfrac{6!}{5!\cdot 1!} \times \dfrac{6!}{1!\cdot 5!}$$
$$=\dfrac{6\cdot 5!}{5!} \times \dfrac{6\cdot 5\cdot 4\cdot 3!}{3\cdot 2\cdot 1\cdot 3!} -\dfrac{6\cdot 5\cdot 4\cdot 3!}{3\cdot 2\cdot 1\cdot 3!} \times \dfrac{6\cdot 5\cdot 4!}{2\cdot 1\cdot 4!} +\dfrac{6\cdot 5!}{5!} \times \dfrac{6\cdot 5!}{5!}$$
$$=6\times 5\times 4-5\times 4\times \dfrac{6\cdot 5}{2} +6\times 6$$
$$=(6\times 5\times 4)-(5\times 4\times 3\times 5)+(6\times 6)$$
$$=120-300+36$$
$$=-144$$
Therefore the coefficient of $$x^{7}$$ is -144.
Hence the correct option is option A.

Note: While solving this type of question you need to know that $${}^{n}C_{r}$$ can be written as, $${}^{n}C_{r}=\dfrac{n!}{r!\cdot \left( n-r\right) !}$$
Where $$n!=n\cdot \left( n-1\right) \cdot \left( n-2\right) \cdots 3\cdot 2\cdot 1$$ and
$$n!=n\cdot \left( n-1\right) !$$
Also while finding the coefficient of $$ ^{7}$$ you no need to multiply each and every term, it will make the solution lengthy.