Question

Find the value of $^{6}{{C}_{0}}^{12}{{C}_{6}}{{-}^{6}}{{C}_{1}}^{11}{{C}_{6}}+\cdots {{+}^{6}}{{C}_{6}}^{6}{{C}_{6}}$.

Answer 1

Hint: Use $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$. Hence calculate the value of each term in the series and hence find the sum of the series. Alternatively, simplify the expression $^{6}{{C}_{r}}^{n-r}{{C}_{6}}$ and hence find an expression in r by putting r = 0,1,2,..6 and solve.

Complete step-by-step answer:
We know that $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
Put n = 6, r =0, we get
$^{6}{{C}_{0}}=\dfrac{6!}{6!}=1$
Put n = 6, r = 1, we get
$^{6}{{C}_{1}}=6$
Put n = 6, r = 2, we get
$^{6}{{C}_{2}}=15$
Put n = 6, r = 3, we get
$^{6}{{C}_{3}}=20$
Similarly $^{6}{{C}_{4}}=15{{,}^{6}}{{C}_{5}}=6$ and $^{6}{{C}_{6}}=1$
Now put n = 12 and r = 6
$^{12}{{C}_{6}}=\dfrac{12!}{6!6!}=924$
Now, we have $\dfrac{^{n-1}{{C}_{r}}}{^{n}{{C}_{r}}}=\dfrac{n-r}{n}$
Hence, we have
$^{n-1}{{C}_{r}}=\dfrac{n-r}{n}{{\times }^{n}}{{C}_{r}}$
Using the above formula, we get
$\begin{align}
  & ^{11}{{C}_{6}}=\dfrac{12-6}{12}{{\times }^{12}}{{C}_{6}}=\dfrac{6}{12}\times 924=462 \\
 & {{\Rightarrow }^{10}}{{C}_{6}}=\dfrac{11-6}{11}{{\times }^{11}}{{C}_{6}}=\dfrac{5}{11}\times 462=210 \\
 & {{\Rightarrow }^{9}}{{C}_{6}}=\dfrac{4}{10}\times 210=84 \\
 & {{\Rightarrow }^{8}}{{C}_{6}}=\dfrac{3}{9}\times 84=28 \\
 & {{\Rightarrow }^{7}}{{C}_{6}}=\dfrac{2}{8}\times 28=7 \\
 & {{\Rightarrow }^{6}}{{C}_{6}}=\dfrac{7}{7}=1 \\
\end{align}$
Hence the given sum becomes
$924-462(6)+210(15)-84(20)+28(15)-7(6)+1(1) = 1$

Note: [1] Alternatively, we have
${{T}_{r}}{{=}^{6}}{{C}_{r}}^{12-r}{{C}_{6}}=\dfrac{6!}{\left( 6-r \right)!r!}\times \dfrac{\left( 12-r \right)!}{\left( 6-r \right)!6!}=\dfrac{\left( 12-r \right)!}{\left( 6-r \right)!\left( 6-r \right)!r!}$
Hence we have $\dfrac{{{T}_{r+1}}}{{{T}_{r}}}=\dfrac{{{\left( 6-r \right)}^{2}}}{\left( 12-r \right)\left( r+1 \right)}$
Now we have ${{T}_{0}}=\dfrac{12!}{6!6!}=924$
Hence, we have
$\begin{align}
  & {{T}_{1}}=924\times \dfrac{{{6}^{2}}}{12}=2772 \\
 & \Rightarrow {{T}_{2}}=2772\times \dfrac{{{5}^{2}}}{\left( 11 \right)\left( 2 \right)}=3150 \\
 & \Rightarrow {{T}_{3}}=3150\times \dfrac{{{4}^{2}}}{10\left( 3 \right)}=1680 \\
 & \Rightarrow {{T}_{4}}=1680\times \dfrac{{{3}^{2}}}{9\left( 4 \right)}=420 \\
 & \Rightarrow {{T}_{5}}=420\times \dfrac{{{2}^{2}}}{\left( 8 \right)\left( 5 \right)}=42 \\
 & \Rightarrow {{T}_{6}}=42\times \dfrac{{{1}^{2}}}{7\left( 6 \right)}=1 \\
\end{align}$
Hence the sum equals $924-2772+3150-1680+420-42+1=1$.
Hence the sum of the series equals 1.
[2] Students often make a mistake by writing $^{12-r}{{C}_{6}}^{6}{{C}_{r}}{{=}^{12-r}}{{C}_{6-r}}^{6}{{C}_{r}}$ and equating it to the coefficient of ${{x}^{6}}$ in the expansion of ${{\left( 1+x \right)}^{6}}{{\left( 1-x \right)}^{12-r}}$, which is not correct and will yield incorrect results. There are many more terms responsible for the coefficient of ${{x}^{6}}$ in the expansion of ${{\left( 1+x \right)}^{6}}{{\left( 1-x \right)}^{12-r}}$ other than $^{12-r}{{C}_{6-r}}^{6}{{C}_{r}}$.