Question

The value of $\left( {^{21}{C_{10}}{ - ^{10}}{C_1}} \right) + \left( {^{21}{C_2}{ - ^{10}}{C_2}} \right) + \left( {^{21}{C_3}{ - ^{10}}{C_3}} \right) + \left( {^{21}{C_4}{ - ^{10}}{C_4}} \right) + ...... + \left( {^{21}{C_{10}}{ - ^{10}}{C_{10}}} \right)$ is:
A. ${2^{21}} - {2^{11}}$
B. ${2^{21}} - {2^{10}}$
C. ${2^{20}} - {2^9}$
D. ${20^{20}} - {2^{10}}$

Answer 1

In the above question, we are given a combination or a grouping of items. Here we will use the formula $C\left( {n,r} \right)$ or $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ . In these types of questions, involving combinations, the order does not matter. $^n{C_r}$ is read as $n$ things taken $r$ at a time. This method is used to calculate the number of possible arrangements in a collection of things.

Formula used:
The formula that will be used to solve this question is: $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ , here $n$ is the number of items and $r$ is the number of items being chosen at a time.

Complete step by step solution:
We have to find the value of $\left( {^{21}{C_{10}}{ - ^{10}}{C_1}} \right) + \left( {^{21}{C_2}{ - ^{10}}{C_2}} \right) + \left( {^{21}{C_3}{ - ^{10}}{C_3}} \right) + \left( {^{21}{C_4}{ - ^{10}}{C_4}} \right) + ...... + \left( {^{21}{C_{10}}{ - ^{10}}{C_{10}}} \right)$ .
To find the value of the above equation, we will first group all the positive and negative terms separately.
$\left( {^{21}{C_1}{ + ^{21}}{C_2} + .....{ + ^{21}}{C_{10}}} \right) - \left( {^{10}{C_1}{ + ^{10}}{C_2} + ...{ + ^{10}}{C_{10}}} \right)$
Now to further solve this we have to use the formula mentioned above $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ on the second bracket. On using this formula, we get,
$\left( {^{21}{C_1}{ + ^{21}}{C_2} + .....{ + ^{21}}{C_{10}}} \right) - \left( {{2^{10}} - 1} \right)$
Now multiply and divide the first bracket by $2$ . We get,
$\dfrac{1}{2}\left( {{2^{21}}{C_1} + {2^{21}}{C_2} + ..... + {2^{21}}{C_{10}}} \right) - \left( {{2^{10}} - 1} \right)$
$\Rightarrow \dfrac{1}{2}\left( {^{21}{C_1}{ + ^{21}}{C_2} + .....{ + ^{21}}{C_{10}} + ........{ + ^{21}}{C_{20}}} \right) - \left( {{2^{10}} - 1} \right)$
Again, using the formula $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ , we get,
$\dfrac{1}{2}\left( {{2^{21}} - 1 - 1} \right) - \left( {{2^{10}} - 1} \right)$
On opening the brackets, we get

$${2^{20}} - 1 - {2^{10}} + 1 \\\ \Rightarrow {2^{20}} - {2^{10}} \\\$$

From this, we derive that our final answer is: ${2^{20}} - {2^{10}}$ .
Hence the correct option is D. ${2^{20}} - {2^{10}}$ .

Note: In order to solve the above question involving combinations, you must always remember the formula used to derive the value of combinations $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ . This formula is used to calculate the number of possible arrangements in a collection of items. All the calculations must be carefully done to avoid any mistakes and calculate the correct answer.