16.0 Multinomial Theorem

Let, $$ \Rightarrow {x_1} + {x_2} + {x_3} + ... + {x_m} = n$$

where, \[\left. \begin{gathered}0 < {x_1} < {a_1} \hspace{1em} \\0 < {x_2} < {a_2} \hspace{1em} \\ \end{gathered} \right\}{\text{ and so on}}\]

Then the number of solutions is the coefficient of ${x^n}$ in the expansion,

$$\begin{equation} \begin{aligned} = ({x^0} + {x^1} + {x^2} + ....{x^{{a_1}}})({x^0} + {x^1} + {x^2} + ....{x^{{a_2}}})....({x^0} + {x^1} + {x^2} + ....{x^{{a_m}}}) \\ \\ = \frac{{(1 - {x^{{a_1} + 1}})(1 - {x^{{a_2} + 1}})..(1 - {x^{{a_m} + 1}})}}{{(1 - x)(1 - x)......(1 - x)}} \\ \\ = \frac{{(1 - {x^{{a_1} + 1}})(1 - {x^{{a_2} + 1}})..(1 - {x^{{a_m} + 1}})}}{{{{(1 - x)}^m}}} \\ \\ = (1 - {x^{{a_1} + 1}})(1 - {x^{{a_2} + 1}})..(1 - {x^{{a_m} + 1}}){(1 - x)^{ - m}} \\ \\ = (1 - {x^{{a_1} + 1}})(1 - {x^{{a_2} + 1}})..(1 - {x^{{a_m} + 1}})(1{ + ^m}{C_1}x{ + ^{m + 1}}{C_2}{x^2}{ + ^{m + 2}}{C_3}{x^3} + .....{ + ^{m + n - 1}}{C_n}{x^n}) \\\end{aligned} \end{equation} $$

Since by binomial theorem, $$ \Rightarrow {(1 - x)^{ - n}} = 1{ + ^n}{C_1}x{ + ^{n + 1}}{C_2}{x^2}{ + ^{n + 2}}{C_3}{x^3} + .....$$

Question 30. Find the number of solution of the equation $a + b + c = 6$, where $a$, $b$, $c$ belong to whole number.

Solution: Number of solutions,

$$\begin{equation} \begin{aligned} = {\text{coefficient of }}{x^6}{\text{ in }}({x^0} + {x^1} + {x^2} + ....{x^6})({x^0} + {x^1} + {x^2} + ....{x^6})({x^0} + {x^1} + {x^2} + ....{x^6}) \\ = {\text{coefficient of }}{x^6}{\text{ in }}\frac{{(1 - {x^7})(1 - {x^7})(1 - {x^7})}}{{(1 - x)(1 - x)(1 - x)}} \\ = {\text{coefficient of }}{x^6}{\text{ in }}{(1 - {x^7})^3}{(1 - x)^{ - 3}} \\ = {\text{coefficient of }}{x^6}{\text{ in }}{(1 - x)^{ - 3}} \\ = {\text{coefficient of }}{x^6}{\text{ in }}\left[ {1{ + ^3}{C_1}x{ + ^{3 + 1}}{C_2}{x^2}{ + ^{3 + 2}}{C_3}{x^3} + .....{ + ^{3 + 6 - 1}}{C_6}{x^6}} \right] \\ = {\,^{3 + 6 - 1}}{C_6} \\ = {\,^8}{C_6} \\ = 28 \\\end{aligned} \end{equation} $$

Question 31. In a candy shop there are four types of candies. In how many ways can a person buy $10$ candies if he decides to take atleast one candyof each variety?

Solution: Let the person select $a$ candies from first variety, $b$ candies from the second, $c$ from the third and $d$ from the fourth.

Then the number of ways of selecting $10$ candies is equal to the number of solution of the equation, $$ \Rightarrow a + b + c + d = 10$$

Given that he takes atleast one from each, he can take only maximum of $7$ from one variety. (i.e. {$1,1,1,7$})

Thus,

$$\begin{equation} \begin{aligned} a = 1,2,...7 \\ b = 1,2,...7 \\ c = 1,2,...7 \\ d = 1,2,...7 \\\end{aligned} \end{equation} $$

So the number of ways, $$ = {\text{coefficient of }}{x^{10}}{\text{ in }}(x + {x^2} + {x^3} + .. + {x^7})(x + {x^2} + {x^3} + .. + {x^7})(x + {x^2} + {x^3} + .. + {x^7})(x + {x^2} + {x^3} + .. + {x^7})$$

Now, lets consider, $$\begin{equation} \begin{aligned} \Rightarrow p = a - 1 \\ \Rightarrow q = b - 1 \\ \Rightarrow r = c - 1 \\ \Rightarrow s = d - 1 \\\end{aligned} \end{equation} $$

Thus,

$$\begin{equation} \begin{aligned} p = 0,1,2,...6 \\ q = 0,1,2,...6 \\ r = 0,1,2,...6 \\ s = 0,1,2,...6 \\\end{aligned} \end{equation} $$

and the equation becomes, $$ \Rightarrow p + 1 + q + 1 + r + 1 + s + 1 = 10$$ i.e. $$ \Rightarrow p + q + r + s = 6$$

Thus, the number of ways,

$$\begin{equation} \begin{aligned} = {\text{coefficient of }}{x^6}{\text{ in }}({x^0} + {x^1} + {x^2} + ....{x^6})({x^0} + {x^1} + {x^2} + ....{x^6})({x^0} + {x^1} + {x^2} + ....{x^6})({x^0} + {x^1} + {x^2} + ....{x^6}) \\ \\ = {\text{coefficient of }}{x^6}{\text{ in }}{x^6}\frac{{(1 - {x^7})(1 - {x^7})(1 - {x^7})(1 - {x^7})}}{{(1 - x)(1 - x)(1 - x)(1 - x)}} \\ \\ = {\text{coefficient of }}{x^6}{\text{ in }}{(1 - {x^7})^4}{(1 - x)^{ - 4}} \\ \\ = {\text{coefficient of }}{x^6}{\text{ in }}{(1 - x)^{ - 4}} \\ \\ = {\text{coefficient of }}{x^6}{\text{ in }} = 1{ + ^4}{C_1}x{ + ^{4 + 1}}{C_2}{x^2}{ + ^{4 + 2}}{C_3}{x^3} + .....{ + ^{4 + 6 - 1}}{C_6}{x^6} \\ \\ = {\,^{4 + 6 - 1}}{C_6} \\ = {\,^9}{C_6} \\ = 84 \\\end{aligned} \end{equation} $$