Binomial Theorem
11.0 Multinomial theorem
11.0 Multinomial theorem
As we know the binomial theorem, $${(x + y)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{x^{n - r}}{y^r}} = \sum\limits_{r = 0}^n {\frac{{n!}}{{r!\left( {n - r} \right)!}}{x^{n - r}}{y^r}} $$
Put $n - r = {r_1},{\text{ }}r = {r_2}$, therefore $${(x + y)^n} = \sum\limits_{{r_1} + {r_2} = n} {\frac{{n!}}{{{r_1}!{r_2}!}}} {x^{{r_1}}}.{y^{{r_2}}}$$
Total number of terms in the expansion of ${(x + y)^n}$ is equal to the number of non-negative integral solution of ${r_1} + {r_2} = n$ i.e., ${}^{n + 2 - 1}{C_{2 - 1}} = {}^{n + 1}{C_1} = n + 1$
In the same way, we can write the multinomial theorem
$${({x_1} + {x_2} + {x_3} + ... + {x_k})^n} = \sum\limits_{{r_1} + {r_2} + ... + {r_k} = n} {\frac{{n!}}{{{r_1}!{r_2}!...{r_k}!}}{x_1}^{{r_1}}.{x_2}^{{r_2}}...{x_k}^{{r_k}}} $$
Here total number of terms in the expansion of ${({x_1} + {x_2} + {x_3} + ... + {x_k})^n}$ is equal to the number of non-negative integral solution of ${{r_1} + {r_2} + ... + {r_k} = n}$ i.e., ${}^{n + k - 1}{C_{k - 1}}$
Question 11. Find the coefficient of ${a^4}{b^3}{c^2}d$ in the expansion of ${(a - b + c - d)^9}$.
Solution: $${(a - b + c - d)^9} = \sum\limits_{{r_1} + {r_2} + {r_3} + {r_4} = 9} {\frac{{9!}}{{{r_1}!{r_2}!{r_3}!{r_4}!}}{{(a)}^{{r_1}}}.{{( - b)}^{{r_2}}}.{{(c)}^{{r_3}}}.} {( - d)^{{r_4}}}$$
We want to get ${a^4}{b^3}{c^2}d$ i.e., ${r_1} = 4,{r_2} = 3,{r_3} = 2,{r_4} = 1$. Therefore, coefficient of ${a^4}{b^3}{c^2}d$ is $${( - 1)^3}( - 1)\frac{{9!}}{{4!3!2!1!}} = 12600$$
