Htykuut

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.....

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
$$sum_{k = 2}^{n - 2} binom{n}{k} = 2^n - 2n - 2$$

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.

As team should consist of at least 2 and at most (n – 2) students out of n students, so required number of ways = 2 students or 3 students ........or (n – 3) students or (n – 2)students.

⇒ number or ways = nC2 + nC3 + nC4 + ...... nCn – 3 + nCn – 2 ...... (I)

nC0 + nC1 + nC2 + nC3 + ........ + nCn – 2 + nCn – 1 + nCn = 2n

⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – nC0 – nC1 – nCn–1 – nCn

⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – 1 – n – n – 1

⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – 2n – 2 ......... (II)

Substituting (II) in (I),

Number of ways = 2n – 2n – 2

Hence, number of ways of forming a team: (1) 2n – 2n – 2

Hope it helps!