Question

If $C_{r}$ stands for ${}_{}^{n}C_{r}^{},$ then the coefficient of $\lambda ^{n}\mu ^{n}$ in the expansion of $\left[\left(1 + \lambda \right) \left(1 + \mu \right) \left(\lambda + \mu \right)\right]^{n}$ is:

• A. $\displaystyle \sum _{r = 0}^{n}C_{r}^{2}$
• B. $\displaystyle \sum _{r = 0}^{n}C_{r + 2}^{2}$
• C. $\displaystyle \sum _{r = 0}^{n}C_{r + 3}^{2}$
• D. $\displaystyle \sum _{r = 0}^{n}C_{r}^{3}$

Answer 1

Answer: (D)

$\left(1 + \lambda \right)^{n}\left(\mu + 1\right)^{n}\left(\lambda + \mu \right)^{n}=\displaystyle \sum _{r = 0}^{n}C_{r}\left(\lambda \right)^{r}\displaystyle \sum _{s = 0}^{n}C_{s}\left(\mu \right)^{n - s}\displaystyle \sum _{t = 0}^{n}C_{t}\left(\lambda \right)^{n - 1}\left(\mu \right)^{t}$
since, $C_{0}^{3}$ is the coefficient of $\lambda ^{0}\mu ^{n - 0}\lambda ^{n - 0}\mu ^{0}$
i.e., $\left(\lambda \right)^{n}\left(\mu \right)^{n}\left(r = s = t = 0\right)$
Now, $C_{1}^{3}$ is the coefficient of $\lambda ^{1}\mu ^{n - 1}\lambda ^{n - 1}\mu ^{1}$
i.e., $\left(\lambda \right)^{n}\left(\mu \right)^{n}\left(r = s = t = 1\right)$ etc.
and $C_{k}^{3}$ is the coefficient of $\lambda ^{k}\mu ^{n - k}\lambda ^{n - k}\mu ^{k}$
i.e., $\left(\lambda \right)^{n}\left(\mu \right)^{n}\left(r = s = t = k\right)$
Hence, the coefficient of $\lambda ^{n}\mu ^{n}$ in
$\left(1 + \lambda \right)^{n}\left(\mu + 1\right)^{n}\left(\lambda + \mu \right)^{n}=C_{0}^{3}+C_{1}^{3}+C_{2}^{3}+\ldots +C_{n}^{3}=\displaystyle \sum _{r = 0}^{n}C_{r}^{3}$