Formation of Teams in Permutation and Combination
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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.....
combinatorics permutations combinations binomial-theorem
edited Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
asked Jul 23 '16 at 8:09
saladi
13010
add a comment |
up vote
0
down vote
favorite
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.....
combinatorics permutations combinations binomial-theorem
edited Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
asked Jul 23 '16 at 8:09
saladi
13010
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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.....
combinatorics permutations combinations binomial-theorem
edited Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
asked Jul 23 '16 at 8:09
saladi
13010
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.....
combinatorics permutations combinations binomial-theorem
combinatorics permutations combinations binomial-theorem
edited Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
asked Jul 23 '16 at 8:09
saladi
13010
edited Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
asked Jul 23 '16 at 8:09
saladi
13010
edited Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
edited Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
edited Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
43.3k93354
asked Jul 23 '16 at 8:09
saladi
13010
asked Jul 23 '16 at 8:09
saladi
13010
asked Jul 23 '16 at 8:09
saladi
13010
13010
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
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
$$sum_{k = 2}^{n - 2} binom{n}{k} = 2^n - 2n - 2$$
answered Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
add a comment |
up vote
2
down vote
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.
answered Jul 23 '16 at 20:20
user84413
22.8k11848
add a comment |
up vote
0
down vote
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!
answered Nov 22 at 19:27
Krishna
1
Please use Latex. Learn latex from here.
– tarit goswami
Nov 22 at 19:32
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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
$$sum_{k = 2}^{n - 2} binom{n}{k} = 2^n - 2n - 2$$
answered Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
add a comment |
up vote
2
down vote
accepted
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
$$sum_{k = 2}^{n - 2} binom{n}{k} = 2^n - 2n - 2$$
answered Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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
$$sum_{k = 2}^{n - 2} binom{n}{k} = 2^n - 2n - 2$$
answered Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
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
$$sum_{k = 2}^{n - 2} binom{n}{k} = 2^n - 2n - 2$$
answered Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
answered Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
answered Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
answered Jul 23 '16 at 8:53
N. F. Taussig
43.3k93354
43.3k93354
add a comment |
add a comment |
up vote
2
down vote
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.
answered Jul 23 '16 at 20:20
user84413
22.8k11848
add a comment |
up vote
2
down vote
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.
answered Jul 23 '16 at 20:20
user84413
22.8k11848
add a comment |
up vote
2
down vote
up vote
2
down vote
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.
answered Jul 23 '16 at 20:20
user84413
22.8k11848
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.
answered Jul 23 '16 at 20:20
user84413
22.8k11848
answered Jul 23 '16 at 20:20
user84413
22.8k11848
answered Jul 23 '16 at 20:20
user84413
22.8k11848
answered Jul 23 '16 at 20:20
user84413
22.8k11848
22.8k11848
add a comment |
add a comment |
up vote
0
down vote
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!
answered Nov 22 at 19:27
Krishna
1
Please use Latex. Learn latex from here.
– tarit goswami
Nov 22 at 19:32
add a comment |
up vote
0
down vote
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!
answered Nov 22 at 19:27
Krishna
1
Please use Latex. Learn latex from here.
– tarit goswami
Nov 22 at 19:32
add a comment |
up vote
0
down vote
up vote
0
down vote
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!
answered Nov 22 at 19:27
Krishna
1
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!
answered Nov 22 at 19:27
Krishna
1
answered Nov 22 at 19:27
Krishna
1
answered Nov 22 at 19:27
Krishna
1
answered Nov 22 at 19:27
Krishna
1
1
Please use Latex. Learn latex from here.
– tarit goswami
Nov 22 at 19:32
add a comment |
Please use Latex. Learn latex from here.
– tarit goswami
Nov 22 at 19:32
Please use Latex. Learn latex from here.
– tarit goswami
Nov 22 at 19:32
Please use Latex. Learn latex from here.
– tarit goswami
Nov 22 at 19:32
add a comment |
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