Question

A class has n students. We have to form a team of the students including at least two students and excluding at least two students. The number of ways of forming the team is:
A. $ {2^n} - 2n $
B. $ {2^n} - 2n - 2 $
C. $ {2^n} - 2n - 4 $
D. $ {2^n} - 2n - 1 $

Answer 1

Hint: The given team must have at least 2 students and exclude at least 2 students, which means that the team can have at most n-2 students. Find the number of ways the team can be formed, in which a team will have at least 2 students and at most n-2 students

Complete step-by-step answer:
We are given that a class has n students and we have to form a team of the students including at least two students and excluding at least two students.
A team can have at least 2 students and exclude at least 2 students, which means at least 2 students from n students must not be included. So it should have at most $ n - 2 $ students. We can select 2 students from n students or 3 students from n or 4 students from n and so on till n-2 students from n students. So, use combinations.
Therefore, the number of ways the team can be formed is $ {}^n{C_2} + {}^n{C_3} + {}^n{C_4} + ..... + {}^n{C_{n - 3}} + {}^n{C_{n - 2}} $
We know that the expansion of $ {\left( {1 + x} \right)^n} $ using binomial expansion is $ \Rightarrow {}^n{C_0}{x^0} + {}^n{C_1}{x^1} + {}^n{C_2}{x^2} + .... + {}^n{C_n}{x^n} $
When the value of x is 1, then
  $
\Rightarrow {\left( {1 + 1} \right)^n} = {}^n{C_0}{1^0} + {}^n{C_1}{1^1} + {}^n{C_2}{1^2} + .... + {}^n{C_n}{1^n} \\
\Rightarrow \to {2^n} = {}^n{C_0}1 + {}^n{C_1}1 + {}^n{C_2}1 + .... + {}^n{C_n}1 \\
  \therefore {2^n} = {}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ..... + {}^n{C_{n - 1}} + {}^n{C_n} \\
  $
So the value of $ {}^n{C_2} + {}^n{C_3} + {}^n{C_4} + ..... + {}^n{C_{n - 3}} + {}^n{C_{n - 2}} $ will be
 $
\Rightarrow {2^n} = {}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ..... + {}^n{C_{n - 1}} + {}^n{C_n} \\
 \Rightarrow {}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ..... + {}^n{C_{n - 1}} + {}^n{C_n} = {2^n} \\
 \Rightarrow {}^n{C_2} + {}^n{C_3} + ..... + {}^n{C_{n - 2}} = {2^n} - \left( {{}^n{C_0} + {}^n{C_1} + {}^n{C_n} + {}^n{C_{n - 1}}} \right) \\
  {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} \\
\Rightarrow {}^n{C_0} = 1,{}^n{C_1} = n,{}^n{C_n} = 1,{}^n{C_{n - 1}} = n \\
 \Rightarrow {}^n{C_2} + {}^n{C_3} + ..... + {}^n{C_{n - 2}} = {2^n} - \left( {1 + n + 1 + n} \right) \\
\Rightarrow {}^n{C_2} + {}^n{C_3} + ..... + {}^n{C_{n - 2}} = {2^n} - 2 - 2n \\
  \therefore {}^n{C_2} + {}^n{C_3} + ..... + {}^n{C_{n - 2}} = {2^n} - 2n - 2 \\
  $
So, the correct answer is “Option B”.

Note: A Permutation is arranging objects in order. Combination is the way of selecting objects from a group of objects. When the order of the objects does not matter it should be considered as Combination and when the order matters it should be considered as Permutation. Do not confuse using a combination instead of a permutation and vice-versa.