Formation of Teams in Permutation and Combination

Question

A class has $n$ students , we have to form a team of the students including at least two and also excluding at least two students. The number of ways of forming the team is

My Approach : To include at least two students the required ways is

C($n$, $2$) $+$ C($n$,$3$) $+$ C($n$,$4$)...........$+$C($n$,$n-2$) But I am not understanding how to calculate the number of ways of excluding at least two students with this..............

Please help.....

Answer 1

Your interpretation that the number of students selected is at least $2$ and at most $n - 2$ is correct, as is your answer. We can use the Binomial Theorem to obtain a closed form.

The Binomial Theorem states that $$(x + y)^n = \sum_{k = 0}^n \binom{n}{k}x^{n - k}y^k$$ We can find $$\sum_{k = 0}^{n} \binom{n}{k}$$ by substituting $1$ for both $x$ and $y$, which yields $$2^n = (1 + 1)^n = \sum_{k = 0}^n \binom{n}{k}1^{n - k}1^k = \sum_{k = 0}^n \binom{n}{k}$$ Comparing this expression with your answer $$\binom{n}{2} + \binom{n}{3} + \cdots + \binom{n}{n - 3} + \binom{n}{n - 2} = \sum_{k = 2}^{n - 2} \binom{n}{k}$$ we have $$\sum_{k = 2}^{n - 2} \binom{n}{k} = \sum_{k = 0}^{n} \binom{n}{k} - \left[\binom{n}{0} + \binom{n}{1} + \binom{n}{n - 1} + \binom{n}{n}\right]$$ Since $$\binom{n}{0} = \binom{n}{n} = 1$$ and $$\binom{n}{1} = \binom{n}{n - 1} = n$$ we obtain $$\sum_{k = 2}^{n - 2} \binom{n}{k} = 2^n - 2n - 2$$

Answer 2

There are $2^n$ possible teams altogether, and there are

$\binom{n}{1}+\binom{n}{n-1}=2n$ teams with $1$ or $n-1$ students and $\binom{n}{0}+\binom{n}{n}=2$ teams with $0$ or $n$ students;

so there are $2^n-2n-2$ teams including at least 2 students and excluding at least 2 students.