ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 5

Saturday, 9 August 2014

CHAPTER 5 - Worked Out Examples 3

Example: 5

How will you expand the multinomial expression ${\left( {{x_1} + {x_2} +\ldots + {x_m}} \right)^n}$ ?

Solution: 5

We will approach this problem using combinatorics. Note that a general term of the expansion would be of the form (without the coefficient)

$x_1^{{n_1}}x_2^{{n_2}}x_3^{n{ _3}}\ldots x_m^{{n_m}}$

$\ldots(1)$

where the various powers must always sum to $n$ (why?).

i.e.,

${n_1} + {n_2} + {n_3} + \ldots + {n_m} = n$

Now, to evaluate the coefficient of the term in $(1)$ , we consider the multinomial expression in expanded form:

$\underbrace {\left( {{x_1} + {x_2} + \ldots + {x_m}} \right)\left( {{x_1} + {x_2} +\ldots + {x_m}} \right) \ldots \left( {{x_1} + {x_2} +\ldots {x_m}} \right)}_{n\,\,{\rm{times}}}$

To generate the term in $(1)$ , we must get ${x_1}$ from ${n_1}$ terms, ${x_2}$ from ${n_1}$ terms and so on. Let us find the number of ways in which this can be done.

First select those ${n_1}$ multinomials that will contribute ${x_1}$ : this can be done in ${}^n{C_{{n_1}}}$ ways. Now, from the remaining $\left( {n - {n_1}} \right)$ multinomials, select those ${n_2}$ multinomials that will contribute ${x_2}$ : this can be done in ${}^{(n - {n_1})}{C_{{n_2}}}$ ways. Continuing this process, we see that the number of ways to get ${x_1}$ from ${n_1}$ ${x_2}$ from ${n_2} \ldots$ and so on, that is, the number of times the term in $(1)$ will be generated in the expansion is

	$^n{C_{{n_1}}} \times {\,^{\left( {n - {n_1}} \right)}}{C_{{n_2}}} \times {\,^{\left( {n - {n_1} - {n_2}} \right)}}{C_{{n_3}}} \times\ldots$
	$= \dfrac{{n!}}{{{n_1}!\left( {n - {n_1}} \right)!}}\,\, \times \,\,\dfrac{{\left( {n - {n_1}} \right)!}}{{{n_2}!\left( {n - {n_1} - {n_2}} \right)!}}\,\, \times \,\,\dfrac{{\left( {n - {n_1} - {n_2}} \right)!}}{{{n_3}!\left( {n - {n_1} - {n_2} - {n_3}} \right)!}}$ $\,\, \times \,\,\ldots$
	$= \,\,\dfrac{{n!}}{{{n_1}!\,\,{n_2}!\ldots {n_m}!}}\,$

This is what is known as the general multinomial coefficient. The multinomial expansion can now be written compactly as

${\left( {{x_1} + {x_2} + \ldots + {x_m}} \right)^n} = \sum {\dfrac{{n!}}{{{n_1}!{n_2}! \ldots {n_m}!}}} \,\,x_1^{{n_1}}x_2^{{n_2}}\ldots \,x_m^{{n_m}}$

where the summation is carried out over all possible combinations of the ${n_i}$ ‘s such that $\sum {{n_i} = n}$ . For example, in ${\left( {{x_1} + {x_2} + {x_3}} \right)^4}$ , let us consider some terms in the expansion:

Multinomial	Term	Coefficient
	${x_1}^2{x_2}{x_3}$	$\dfrac{{4!}}{{2!1!1!}} = 12$
	${x_1}^3{x_2}$	$\dfrac{{4!}}{{3!1!}} = 4$
${\left( {{x_1} + {x_2} + {x_3}} \right)^4}$	${x_2}^4$	$\dfrac{{4!}}{{4!}} = 1$
	${x_1}{x_2}{x_3}^2$	$\dfrac{{4!}}{{1!1!2!}} = 12$
	$\vdots$

Example: 6

Find the coefficient of ${x^4}$ in the expansion of ${\left( {1 + x - \dfrac{2}{{{x^2}}}} \right)^{10}}$ .

Solution: 6

From the previous example, the general term in the expansion will be

	$\dfrac{{10!}}{{{n_1}!\;{n_2}!\;{n_3}!}}\;{(1)^{{n_1}}}{(x)^{{n_{^2}}}}{\left( {\dfrac{{ - 2}}{{{x^2}}}} \right)^{{n_{^3}}}}$
	$= \dfrac{{10}}{{{n_1}!\;{n_2}!\;{n_3}!}}\;{x^{{n_2} - 2{n_3}}}{( - 2)^{{n_3}}}$