9.0 Properties of binomial coefficient

In the expansion of ${(a + x)^n}$, put $a=1$ we get,

$${(1 + x)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ... + {}^n{C_n}{x^n}$$

1. If $x=1$, then $$\begin{equation} \begin{aligned} {(1 + 1)^n} = {}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ... + {}^n{C_n} \\ {2^n} = {}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ... + {}^n{C_n} \\\end{aligned} \end{equation} $$

or, $$\sum\limits_{r = 0}^n {{}^n{C_r} = {2^n}} $$

i.e., Sum of binomial coefficients is ${2^n}$.

2. If $x=-1$, then

$$\begin{equation} \begin{aligned} {(1 - 1)^n}{ = ^n}{C_0}{ - ^n}{C_1}{ + ^n}{C_2} - ...{ + ^n}{C_n}{( - 1)^n} \\ 0{ = ^n}{C_0}{ - ^n}{C_1}{ + ^n}{C_2} - ...{ + ^n}{C_n}{( - 1)^n} \\ {\text{or,}} \\ \sum\limits_{r = 0}^n {{{( - 1)}^r}{.^n}{C_r} = 0} \\\end{aligned} \end{equation} $$

Now, we can write the above equation as $$^n{C_1}{ + ^n}{C_3}{ + ^n}{C_5}...{ = ^n}{C_0}{ + ^n}{C_2}{ + ^n}{C_4} + ...$$

i.e., Odd binomial coefficients = Even binomial coefficients=$x$.

Now, Sum of odd binomial coefficients and even binomial coefficients = Sum of binomial coefficients

i.e., $$\begin{equation} \begin{aligned} x + x = {2^n} \\ 2x = {2^n} \\ x = {2^{n - 1}} \\\end{aligned} \end{equation} $$

3. Sum of two consecutive binomial coefficients i.e., $${}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}$$

Proof: Take L.H.S.

$$\begin{equation} \begin{aligned} ^n{C_r}{ + ^n}{C_{r - 1}} = \frac{{n!}}{{r!\left( {n - r} \right)!}} + \frac{{n!}}{{(r - 1)!\left( {n - r + 1} \right)!}} \\ {\text{ = }}\frac{{n!}}{{(r - 1)!\left( {n - r} \right)!}}\left[ {\frac{1}{r} + \frac{1}{{n - r + 1}}} \right] \\ {\text{ = }}\frac{{n!}}{{(r - 1)!\left( {n - r} \right)!}}\frac{{(n + 1)}}{{r(n - r + 1)}} \\ {\text{ = }}\frac{{(n + 1)!}}{{r!\left( {n - r + 1} \right)!}} = {}^{n + 1}{C_r} = R.H.S. \\\end{aligned} \end{equation} $$

4. Ratio of two consecutive binomial coefficients i.e., $$\frac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \frac{{n - r + 1}}{r}$$

5. $${}^n{C_r} = \frac{n}{r}{}^{n - 1}{C_{r - 1}} = \frac{{n(n - 1)}}{{r(r - 1)}}{}^{n - 2}{C_{r - 2}} = ... = \frac{{n(n - 1)(n - 2)...(n - (r - 1))}}{{r(r - 1)(r - 2)...2.1}}{}^{n - 2}{C_{r - 2}}$$