Mixing
Find the number of ways in which 6 boys and 6 girls can be seated in a row so that all the girls are never...
$begingroup$
Find the number of ways in which 6 boys and 6 girls can be seated in a row so that all the girls are never together.
Attempt:
total number of ways - number of ways in which all girls are together
$= 12! - 2!times(6!)times (6!)$
But answer given is $12! - 7!6!$
Is the given answer wrong?
combinatorics
asked Oct 25 '18 at 12:04
AbcdAbcd
3,05531237
$endgroup$
add a comment |
$begingroup$
Find the number of ways in which 6 boys and 6 girls can be seated in a row so that all the girls are never together.
Attempt:
total number of ways - number of ways in which all girls are together
$= 12! - 2!times(6!)times (6!)$
But answer given is $12! - 7!6!$
Is the given answer wrong?
combinatorics
asked Oct 25 '18 at 12:04
AbcdAbcd
3,05531237
$endgroup$
add a comment |
$begingroup$
Find the number of ways in which 6 boys and 6 girls can be seated in a row so that all the girls are never together.
Attempt:
total number of ways - number of ways in which all girls are together
$= 12! - 2!times(6!)times (6!)$
But answer given is $12! - 7!6!$
Is the given answer wrong?
combinatorics
asked Oct 25 '18 at 12:04
AbcdAbcd
3,05531237
$endgroup$
Find the number of ways in which 6 boys and 6 girls can be seated in a row so that all the girls are never together.
Attempt:
total number of ways - number of ways in which all girls are together
$= 12! - 2!times(6!)times (6!)$
But answer given is $12! - 7!6!$
Is the given answer wrong?
combinatorics
combinatorics
asked Oct 25 '18 at 12:04
AbcdAbcd
3,05531237
asked Oct 25 '18 at 12:04
AbcdAbcd
3,05531237
asked Oct 25 '18 at 12:04
AbcdAbcd
3,05531237
asked Oct 25 '18 at 12:04
AbcdAbcd
3,05531237
asked Oct 25 '18 at 12:04
AbcdAbcd
3,05531237
3,05531237
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
With any constraint the number of possible combination is $12!$
If all the girls are together, we can think the set to be of $6+1$ members which can be arranged in $7!$ ways
and again the $6$ girls can be arranged in $6!$ ways
answered Oct 25 '18 at 12:09
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Reminds me of math.stackexchange.com/questions/697433/…
$endgroup$
– lab bhattacharjee
Oct 25 '18 at 12:47
add a comment |
$begingroup$
All the girls together means a sequence of 6 girls in a row.
This sequence can start at position 1, 2, 3, 4, 5, 6 and 7.
There are $6!$ ways to place the girls and $6!$ ways to place the boys.
So you have $7 cdot 6! cdot 6! = 7! cdot 6!$ possible ways to violate the rule.
Thus there are $12! - 7! cdot 6!$ ways that respect the rule.
answered Oct 25 '18 at 12:10
RonaldRonald
673510
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
With any constraint the number of possible combination is $12!$
If all the girls are together, we can think the set to be of $6+1$ members which can be arranged in $7!$ ways
and again the $6$ girls can be arranged in $6!$ ways
answered Oct 25 '18 at 12:09
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Reminds me of math.stackexchange.com/questions/697433/…
$endgroup$
– lab bhattacharjee
Oct 25 '18 at 12:47
add a comment |
$begingroup$
With any constraint the number of possible combination is $12!$
If all the girls are together, we can think the set to be of $6+1$ members which can be arranged in $7!$ ways
and again the $6$ girls can be arranged in $6!$ ways
answered Oct 25 '18 at 12:09
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Reminds me of math.stackexchange.com/questions/697433/…
$endgroup$
– lab bhattacharjee
Oct 25 '18 at 12:47
add a comment |
$begingroup$
With any constraint the number of possible combination is $12!$
If all the girls are together, we can think the set to be of $6+1$ members which can be arranged in $7!$ ways
and again the $6$ girls can be arranged in $6!$ ways
answered Oct 25 '18 at 12:09
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
With any constraint the number of possible combination is $12!$
If all the girls are together, we can think the set to be of $6+1$ members which can be arranged in $7!$ ways
and again the $6$ girls can be arranged in $6!$ ways
answered Oct 25 '18 at 12:09
lab bhattacharjeelab bhattacharjee
226k15157275
answered Oct 25 '18 at 12:09
lab bhattacharjeelab bhattacharjee
226k15157275
answered Oct 25 '18 at 12:09
lab bhattacharjeelab bhattacharjee
226k15157275
answered Oct 25 '18 at 12:09
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
$begingroup$
Reminds me of math.stackexchange.com/questions/697433/…
$endgroup$
– lab bhattacharjee
Oct 25 '18 at 12:47
add a comment |
$begingroup$
Reminds me of math.stackexchange.com/questions/697433/…
$endgroup$
– lab bhattacharjee
Oct 25 '18 at 12:47
$begingroup$
Reminds me of math.stackexchange.com/questions/697433/…
$endgroup$
– lab bhattacharjee
Oct 25 '18 at 12:47
$begingroup$
Reminds me of math.stackexchange.com/questions/697433/…
$endgroup$
– lab bhattacharjee
Oct 25 '18 at 12:47
add a comment |
$begingroup$
All the girls together means a sequence of 6 girls in a row.
This sequence can start at position 1, 2, 3, 4, 5, 6 and 7.
There are $6!$ ways to place the girls and $6!$ ways to place the boys.
So you have $7 cdot 6! cdot 6! = 7! cdot 6!$ possible ways to violate the rule.
Thus there are $12! - 7! cdot 6!$ ways that respect the rule.
answered Oct 25 '18 at 12:10
RonaldRonald
673510
$endgroup$
add a comment |
$begingroup$
All the girls together means a sequence of 6 girls in a row.
This sequence can start at position 1, 2, 3, 4, 5, 6 and 7.
There are $6!$ ways to place the girls and $6!$ ways to place the boys.
So you have $7 cdot 6! cdot 6! = 7! cdot 6!$ possible ways to violate the rule.
Thus there are $12! - 7! cdot 6!$ ways that respect the rule.
answered Oct 25 '18 at 12:10
RonaldRonald
673510
$endgroup$
add a comment |
$begingroup$
All the girls together means a sequence of 6 girls in a row.
This sequence can start at position 1, 2, 3, 4, 5, 6 and 7.
There are $6!$ ways to place the girls and $6!$ ways to place the boys.
So you have $7 cdot 6! cdot 6! = 7! cdot 6!$ possible ways to violate the rule.
Thus there are $12! - 7! cdot 6!$ ways that respect the rule.
answered Oct 25 '18 at 12:10
RonaldRonald
673510
$endgroup$
All the girls together means a sequence of 6 girls in a row.
This sequence can start at position 1, 2, 3, 4, 5, 6 and 7.
There are $6!$ ways to place the girls and $6!$ ways to place the boys.
So you have $7 cdot 6! cdot 6! = 7! cdot 6!$ possible ways to violate the rule.
Thus there are $12! - 7! cdot 6!$ ways that respect the rule.
answered Oct 25 '18 at 12:10
RonaldRonald
673510
answered Oct 25 '18 at 12:10
RonaldRonald
673510
answered Oct 25 '18 at 12:10
RonaldRonald
673510
answered Oct 25 '18 at 12:10
RonaldRonald
673510
673510
add a comment |
add a comment |
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