Permutations and Combinations
8.0 Greatest Term
8.0 Greatest Term
$${ = ^n}{C_0}{ + ^n}{C_1}{ + ^n}{C_2} + .........{ + ^n}{C_r}$$
For a term to be greatest, it should be greater than the preceeding and succeeding terms.
Thus, $$^n{C_r}{ > ^n}{C_{r - 1}}\quad {\text{and}}{\quad ^n}{C_r}{ > ^n}{C_{r + 1}}$$
(A) $^n{C_r}{ > ^n}{C_{r - 1}}$
$$\begin{equation} \begin{aligned} \Rightarrow \frac{{n!}}{{r!\left( {n - r} \right)!}} > \frac{{n!}}{{(r - 1)!\left( {n - r + 1} \right)!}} \\ \Rightarrow \frac{1}{r} > \frac{1}{{\left( {n - r + 1} \right)}} \\ \Rightarrow n - r + 1 > r \\ \Rightarrow n + 1 > 2r \\ \Rightarrow \frac{{n + 1}}{2} > r \\\end{aligned} \end{equation} $$
(B) $^n{C_r}{ > ^n}{C_{r + 1}}$
$$\begin{equation} \begin{aligned} \Rightarrow \frac{{n!}}{{r!\left( {n - r} \right)!}} > \frac{{n!}}{{(r + 1)!\left( {n - r - 1} \right)!}} \\ \Rightarrow \frac{{n!}}{{r!\left( {n - r} \right)(n - r - 1)!}} > \frac{{n!}}{{(r + 1)(r!)\left( {n - r - 1} \right)!}} \\ \Rightarrow \frac{1}{{\left( {n - r} \right)}} > \frac{1}{{(r + 1)}} \\ \Rightarrow r + 1 > n - r \\ \Rightarrow 2r > n - 1 \\ \Rightarrow \frac{{n - 1}}{2} < r \\\end{aligned} \end{equation} $$
Thus , $$\frac{{n - 1}}{2} < r < \frac{{n + 1}}{2}$$
(i) For even,
$r$ must be a whole number. Thus $r$ is, $$r = \frac{n}{2}$$
(ii) For odd,
$$r = \frac{{n - 1}}{2}\quad {\text{or}}\quad r = \frac{{n + 1}}{2}$$
Question 21. Find the greatest binomial coefficient in the expansion: (a) ${(1 + x)^{11}}$ (b) ${(1 + x)^{14}}$
Solution:
(a) Here, $n = 11$, i.e. $n$ is an odd number.
Thus the greatest binomial coefficients are, $$\begin{equation} \begin{aligned} = {\,^{11}}{C_{\frac{{(11 - 1)}}{2}}} \\ = {\,^{11}}{C_5} \\\end{aligned} \end{equation} $$ and $$\begin{equation} \begin{aligned} = {\,^{11}}{C_{\frac{{(11 + 1)}}{2}}} \\ = {\,^{11}}{C_6} \\\end{aligned} \end{equation} $$
(b) Here $n = 14$, i.e. $n$ is an even number.
Thus the greatest binomial coefficient is, $$\begin{equation} \begin{aligned} = {\,^{14}}{C_{\frac{{14}}{2}}} \\ = {\,^{14}}{C_7} \\\end{aligned} \end{equation} $$
