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

Hint: 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.