In how many difference ways can six players be arranged in a line such that two of them, Abhinav and Manjesh are never together?

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here is my attempt

since we have six persons so the total number of way of arranging six persons in a line is $6!$

now since 'Abhinav' and 'Manjesh' is saying never together so we can subtract the total number of way of arranging 'Abhinav' and 'Manjesh' together.
so the total number of ways we can do this is $6! - 5!2! = 480.$

but the actual answer is $frac 6!2!= 360.$

can anybody advise on this question where I am wrong?
any effort is appreciatable.

edited Aug 27 at 7:37

Entrepreneur

339111

asked Aug 27 at 6:49

Dharmendra Parmar

897

The answer seems to be $480$
â€“Â Entrepreneur
Aug 27 at 7:10

If the actual answer is 360 -- what is the actual question? Perhaps the question as asked is slightly different from the question as you described. Or -- the 360 is simply mistaken.
â€“Â John Coleman
Aug 27 at 12:06

add a commentÂ |Â

up vote
4
down vote

favorite

here is my attempt

but the actual answer is $frac 6!2!= 360.$

can anybody advise on this question where I am wrong?
any effort is appreciatable.

edited Aug 27 at 7:37

Entrepreneur

339111

asked Aug 27 at 6:49

Dharmendra Parmar

897

The answer seems to be $480$
â€“Â Entrepreneur
Aug 27 at 7:10

If the actual answer is 360 -- what is the actual question? Perhaps the question as asked is slightly different from the question as you described. Or -- the 360 is simply mistaken.
â€“Â John Coleman
Aug 27 at 12:06

add a commentÂ |Â

up vote
4
down vote

favorite

here is my attempt

but the actual answer is $frac 6!2!= 360.$

can anybody advise on this question where I am wrong?
any effort is appreciatable.

edited Aug 27 at 7:37

Entrepreneur

339111

asked Aug 27 at 6:49

Dharmendra Parmar

897

here is my attempt

but the actual answer is $frac 6!2!= 360.$

can anybody advise on this question where I am wrong?
any effort is appreciatable.

edited Aug 27 at 7:37

Entrepreneur

339111

asked Aug 27 at 6:49

Dharmendra Parmar

897

edited Aug 27 at 7:37

Entrepreneur

339111

edited Aug 27 at 7:37

Entrepreneur

339111

edited Aug 27 at 7:37

Entrepreneur

339111

asked Aug 27 at 6:49

Dharmendra Parmar

897

asked Aug 27 at 6:49

Dharmendra Parmar

897

asked Aug 27 at 6:49

Dharmendra Parmar

897

The answer seems to be $480$
â€“Â Entrepreneur
Aug 27 at 7:10

If the actual answer is 360 -- what is the actual question? Perhaps the question as asked is slightly different from the question as you described. Or -- the 360 is simply mistaken.
â€“Â John Coleman
Aug 27 at 12:06

add a commentÂ |Â

The answer seems to be $480$
â€“Â Entrepreneur
Aug 27 at 7:10

If the actual answer is 360 -- what is the actual question? Perhaps the question as asked is slightly different from the question as you described. Or -- the 360 is simply mistaken.
â€“Â John Coleman
Aug 27 at 12:06

The answer seems to be $480$
â€“Â Entrepreneur
Aug 27 at 7:10

If the actual answer is 360 -- what is the actual question? Perhaps the question as asked is slightly different from the question as you described. Or -- the 360 is simply mistaken.
â€“Â John Coleman
Aug 27 at 12:06

add a commentÂ |Â

4 Answers
4

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up vote
4
down vote

Alternatively:
$$A _ _ _ _ _ Rightarrow 4cdot 4!\
_ A _ _ _ _ Rightarrow 3cdot 4!\
_ _ A _ _ _ Rightarrow 3cdot 4!\
_ _ _ A _ _ Rightarrow 3cdot 4!\
_ _ _ _ A _ Rightarrow 3cdot 4!\
_ _ _ _ _ A Rightarrow 4cdot 4!\$$
Hence:
$$2cdot (4cdot 4!+3cdot 4!+3cdot 4!)=20cdot 24=480.$$

answered Aug 27 at 7:33

farruhota

14.9k2633

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up vote
3
down vote

I agree with your solution, indeed let consider

the overall permutations $6!$

the configurations to be excluded $2cdot 5 cdot 4!$

that is

$$6!-2cdot 5 cdot 4!=720-240=480$$

We can also derive the result as follow

we can arrange A&M such that they never are together in $2cdot (4+3+2+1)=20$ ways

for each configuration we can arrange the others in $4!=24$ ways

therefore

$$20cdot 24 =480$$

edited Aug 27 at 7:26

answered Aug 27 at 7:16

gimusi

70.7k73786

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up vote
3
down vote

I got one more way to answer this question

we have to arrange six people in a line where A&M are never together so what we will do is we will separate A&M alone so we will rest with 4 people

 - - - -

assure these are the 4 places where 4 people have to be arranged so this could be done in 4!.
now we will use the gap method since we have 4 people so there would be 5 gaps in between 4 people now we can select 2 gaps out of 5 gaps at a time to allocate to A&M respectively and this could be done in 5C2 2!

so the total number of ways would become

4! 5C2 2! = 480

edited Aug 27 at 9:52

answered Aug 27 at 8:04

Dharmendra Parmar

897

add a commentÂ |Â

up vote
1
down vote

If you know how to program, you could brute-force it as a sanity check. A Python example:

from itertools import permutations

players = ['A','B','C','D','E','F']

valid_perms = [p for p in permutations(players) if abs(p.index('A') - p.index('B')) > 1]
print(len(valid_perms)) #prints 480

answered Aug 27 at 12:18

John Coleman

3,60811122

add a commentÂ |Â

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
4
down vote

answered Aug 27 at 7:33

farruhota

14.9k2633

add a commentÂ |Â

up vote
4
down vote

answered Aug 27 at 7:33

farruhota

14.9k2633

add a commentÂ |Â

up vote
4
down vote

answered Aug 27 at 7:33

farruhota

14.9k2633

answered Aug 27 at 7:33

farruhota

14.9k2633

answered Aug 27 at 7:33

farruhota

14.9k2633

answered Aug 27 at 7:33

farruhota

14.9k2633

answered Aug 27 at 7:33

farruhota

14.9k2633

add a commentÂ |Â

up vote
3
down vote

I agree with your solution, indeed let consider

the overall permutations $6!$

the configurations to be excluded $2cdot 5 cdot 4!$

that is

$$6!-2cdot 5 cdot 4!=720-240=480$$

We can also derive the result as follow

we can arrange A&M such that they never are together in $2cdot (4+3+2+1)=20$ ways

for each configuration we can arrange the others in $4!=24$ ways

therefore

$$20cdot 24 =480$$

edited Aug 27 at 7:26

answered Aug 27 at 7:16

gimusi

70.7k73786

add a commentÂ |Â

up vote
3
down vote

I agree with your solution, indeed let consider

the overall permutations $6!$

the configurations to be excluded $2cdot 5 cdot 4!$

that is

$$6!-2cdot 5 cdot 4!=720-240=480$$

We can also derive the result as follow

we can arrange A&M such that they never are together in $2cdot (4+3+2+1)=20$ ways

for each configuration we can arrange the others in $4!=24$ ways

therefore

$$20cdot 24 =480$$

edited Aug 27 at 7:26

answered Aug 27 at 7:16

gimusi

70.7k73786

add a commentÂ |Â

up vote
3
down vote

I agree with your solution, indeed let consider

the overall permutations $6!$

the configurations to be excluded $2cdot 5 cdot 4!$

that is

$$6!-2cdot 5 cdot 4!=720-240=480$$

We can also derive the result as follow

we can arrange A&M such that they never are together in $2cdot (4+3+2+1)=20$ ways

for each configuration we can arrange the others in $4!=24$ ways

therefore

$$20cdot 24 =480$$

edited Aug 27 at 7:26

answered Aug 27 at 7:16

gimusi

70.7k73786

I agree with your solution, indeed let consider

the overall permutations $6!$

the configurations to be excluded $2cdot 5 cdot 4!$

that is

$$6!-2cdot 5 cdot 4!=720-240=480$$

We can also derive the result as follow

we can arrange A&M such that they never are together in $2cdot (4+3+2+1)=20$ ways

for each configuration we can arrange the others in $4!=24$ ways

therefore

$$20cdot 24 =480$$

edited Aug 27 at 7:26

answered Aug 27 at 7:16

gimusi

70.7k73786

edited Aug 27 at 7:26

answered Aug 27 at 7:16

gimusi

70.7k73786

answered Aug 27 at 7:16

gimusi

70.7k73786

answered Aug 27 at 7:16

gimusi

70.7k73786

add a commentÂ |Â

up vote
3
down vote

I got one more way to answer this question

we have to arrange six people in a line where A&M are never together so what we will do is we will separate A&M alone so we will rest with 4 people

 - - - -

so the total number of ways would become

4! 5C2 2! = 480

edited Aug 27 at 9:52

answered Aug 27 at 8:04

Dharmendra Parmar

897

add a commentÂ |Â

up vote
3
down vote

I got one more way to answer this question

we have to arrange six people in a line where A&M are never together so what we will do is we will separate A&M alone so we will rest with 4 people

 - - - -

so the total number of ways would become

4! 5C2 2! = 480

edited Aug 27 at 9:52

answered Aug 27 at 8:04

Dharmendra Parmar

897

add a commentÂ |Â

up vote
3
down vote

I got one more way to answer this question

we have to arrange six people in a line where A&M are never together so what we will do is we will separate A&M alone so we will rest with 4 people

 - - - -

so the total number of ways would become

4! 5C2 2! = 480

edited Aug 27 at 9:52

answered Aug 27 at 8:04

Dharmendra Parmar

897

I got one more way to answer this question

we have to arrange six people in a line where A&M are never together so what we will do is we will separate A&M alone so we will rest with 4 people

 - - - -

so the total number of ways would become

4! 5C2 2! = 480

edited Aug 27 at 9:52

answered Aug 27 at 8:04

Dharmendra Parmar

897

edited Aug 27 at 9:52

answered Aug 27 at 8:04

Dharmendra Parmar

897

answered Aug 27 at 8:04

Dharmendra Parmar

897

answered Aug 27 at 8:04

Dharmendra Parmar

897

add a commentÂ |Â

up vote
1
down vote

If you know how to program, you could brute-force it as a sanity check. A Python example:

from itertools import permutations

players = ['A','B','C','D','E','F']

valid_perms = [p for p in permutations(players) if abs(p.index('A') - p.index('B')) > 1]
print(len(valid_perms)) #prints 480

answered Aug 27 at 12:18

John Coleman

3,60811122

add a commentÂ |Â

up vote
1
down vote

If you know how to program, you could brute-force it as a sanity check. A Python example:

from itertools import permutations

players = ['A','B','C','D','E','F']

valid_perms = [p for p in permutations(players) if abs(p.index('A') - p.index('B')) > 1]
print(len(valid_perms)) #prints 480

answered Aug 27 at 12:18

John Coleman

3,60811122

add a commentÂ |Â

up vote
1
down vote

If you know how to program, you could brute-force it as a sanity check. A Python example:

from itertools import permutations

players = ['A','B','C','D','E','F']

valid_perms = [p for p in permutations(players) if abs(p.index('A') - p.index('B')) > 1]
print(len(valid_perms)) #prints 480

answered Aug 27 at 12:18

John Coleman

3,60811122

If you know how to program, you could brute-force it as a sanity check. A Python example:

from itertools import permutations

players = ['A','B','C','D','E','F']

valid_perms = [p for p in permutations(players) if abs(p.index('A') - p.index('B')) > 1]
print(len(valid_perms)) #prints 480

answered Aug 27 at 12:18

John Coleman

3,60811122

answered Aug 27 at 12:18

John Coleman

3,60811122

answered Aug 27 at 12:18

John Coleman

3,60811122

answered Aug 27 at 12:18

John Coleman

3,60811122

add a commentÂ |Â

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