3 girls and 4 boys were standing in a circle . What is the probability that two girls are together but one is...
Question:
3 girls and 4 boys were standing in a circle . What is the probability that two girls are together but one is not with them ?
combinatorics recreational-mathematics
asked Dec 6 '15 at 2:33
Abu Bardewa
84213
add a comment |
Question:
3 girls and 4 boys were standing in a circle . What is the probability that two girls are together but one is not with them ?
combinatorics recreational-mathematics
asked Dec 6 '15 at 2:33
Abu Bardewa
84213
add a comment |
Question:
3 girls and 4 boys were standing in a circle . What is the probability that two girls are together but one is not with them ?
combinatorics recreational-mathematics
asked Dec 6 '15 at 2:33
Abu Bardewa
84213
Question:
3 girls and 4 boys were standing in a circle . What is the probability that two girls are together but one is not with them ?
combinatorics recreational-mathematics
combinatorics recreational-mathematics
asked Dec 6 '15 at 2:33
Abu Bardewa
84213
asked Dec 6 '15 at 2:33
Abu Bardewa
84213
asked Dec 6 '15 at 2:33
Abu Bardewa
84213
asked Dec 6 '15 at 2:33
Abu Bardewa
84213
asked Dec 6 '15 at 2:33
Abu Bardewa
84213
84213
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Since only probability has been asked for, there is a very simple way.
Ignore the boys, and seat any two girls together.
The $3rd$ girl has $3$ permissible places out of 5,
thus $Pr = dfrac35 = 0.6$
If you insist on doing it in a conventional manner:
Unless otherwise specified, seats in a circle are taken as unnumbered, and formula is $(n-1)!$
Thus total ways of seating = $6!$
For the favorable ways :-
$binom32 = 3$ ways of choosing the two "together" girls,
$2$ ways of seating them somewhere $(A-B / B-A)$,
$3$ ways of seating the remaining girl non-adjacently,
$4!$ ways of seating the boys,
Putting everything together, $Pr = dfrac{3cdot2cdot3cdot4!}{6!} = 0.6$
answered Dec 6 '15 at 5:16
true blue anil
20.5k11841
I see that the people are standing, not sitting. Not that it matters !
– true blue anil
Dec 6 '15 at 6:56
add a comment |
a bit off topic but are you applying to Oxbridge?
And now to the question;
The total # ways of arranging is$frac{7!}{7}$ (due to rotations)
The # of ways of putting two girls together is $3choose 2$
Now they must have boys on either side so $4 choose 2$
And finally the other 3 can be arranged in $3! ways$
Pick two adjacent slots from your 7 $6choose 1$
then put the two girls in and multiply by 2 as they can be arranged in 2 ways
Then select the adjacent boys and again multiply by 2 and finally arrange the last.(account for rotations) This gives;
$$frac{{6choose 1}cdot {3choose 2}cdot 2 cdot {4choose 2}cdot 2 cdot 3!}{6}$$
ways and hence the probability is the second expression divided by the first.
answered Dec 6 '15 at 2:44
KingJ
511212
I think you should review your solution. The answer of $0.3$ it yields doesn't seem correct.
– true blue anil
Dec 6 '15 at 6:49
Ah yes, forgot to include the second 2 should be fixed now
– KingJ
Dec 6 '15 at 18:29
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Since only probability has been asked for, there is a very simple way.
Ignore the boys, and seat any two girls together.
The $3rd$ girl has $3$ permissible places out of 5,
thus $Pr = dfrac35 = 0.6$
If you insist on doing it in a conventional manner:
Unless otherwise specified, seats in a circle are taken as unnumbered, and formula is $(n-1)!$
Thus total ways of seating = $6!$
For the favorable ways :-
$binom32 = 3$ ways of choosing the two "together" girls,
$2$ ways of seating them somewhere $(A-B / B-A)$,
$3$ ways of seating the remaining girl non-adjacently,
$4!$ ways of seating the boys,
Putting everything together, $Pr = dfrac{3cdot2cdot3cdot4!}{6!} = 0.6$
answered Dec 6 '15 at 5:16
true blue anil
20.5k11841
I see that the people are standing, not sitting. Not that it matters !
– true blue anil
Dec 6 '15 at 6:56
add a comment |
Since only probability has been asked for, there is a very simple way.
Ignore the boys, and seat any two girls together.
The $3rd$ girl has $3$ permissible places out of 5,
thus $Pr = dfrac35 = 0.6$
If you insist on doing it in a conventional manner:
Unless otherwise specified, seats in a circle are taken as unnumbered, and formula is $(n-1)!$
Thus total ways of seating = $6!$
For the favorable ways :-
$binom32 = 3$ ways of choosing the two "together" girls,
$2$ ways of seating them somewhere $(A-B / B-A)$,
$3$ ways of seating the remaining girl non-adjacently,
$4!$ ways of seating the boys,
Putting everything together, $Pr = dfrac{3cdot2cdot3cdot4!}{6!} = 0.6$
answered Dec 6 '15 at 5:16
true blue anil
20.5k11841
I see that the people are standing, not sitting. Not that it matters !
– true blue anil
Dec 6 '15 at 6:56
add a comment |
Since only probability has been asked for, there is a very simple way.
Ignore the boys, and seat any two girls together.
The $3rd$ girl has $3$ permissible places out of 5,
thus $Pr = dfrac35 = 0.6$
If you insist on doing it in a conventional manner:
Unless otherwise specified, seats in a circle are taken as unnumbered, and formula is $(n-1)!$
Thus total ways of seating = $6!$
For the favorable ways :-
$binom32 = 3$ ways of choosing the two "together" girls,
$2$ ways of seating them somewhere $(A-B / B-A)$,
$3$ ways of seating the remaining girl non-adjacently,
$4!$ ways of seating the boys,
Putting everything together, $Pr = dfrac{3cdot2cdot3cdot4!}{6!} = 0.6$
answered Dec 6 '15 at 5:16
true blue anil
20.5k11841
Since only probability has been asked for, there is a very simple way.
Ignore the boys, and seat any two girls together.
The $3rd$ girl has $3$ permissible places out of 5,
thus $Pr = dfrac35 = 0.6$
If you insist on doing it in a conventional manner:
Unless otherwise specified, seats in a circle are taken as unnumbered, and formula is $(n-1)!$
Thus total ways of seating = $6!$
For the favorable ways :-
$binom32 = 3$ ways of choosing the two "together" girls,
$2$ ways of seating them somewhere $(A-B / B-A)$,
$3$ ways of seating the remaining girl non-adjacently,
$4!$ ways of seating the boys,
Putting everything together, $Pr = dfrac{3cdot2cdot3cdot4!}{6!} = 0.6$
answered Dec 6 '15 at 5:16
true blue anil
20.5k11841
edited Dec 6 '15 at 5:39
answered Dec 6 '15 at 5:16
true blue anil
20.5k11841
answered Dec 6 '15 at 5:16
true blue anil
20.5k11841
answered Dec 6 '15 at 5:16
true blue anil
20.5k11841
20.5k11841
I see that the people are standing, not sitting. Not that it matters !
– true blue anil
Dec 6 '15 at 6:56
add a comment |
I see that the people are standing, not sitting. Not that it matters !
– true blue anil
Dec 6 '15 at 6:56
I see that the people are standing, not sitting. Not that it matters !
– true blue anil
Dec 6 '15 at 6:56
I see that the people are standing, not sitting. Not that it matters !
– true blue anil
Dec 6 '15 at 6:56
add a comment |
a bit off topic but are you applying to Oxbridge?
And now to the question;
The total # ways of arranging is$frac{7!}{7}$ (due to rotations)
The # of ways of putting two girls together is $3choose 2$
Now they must have boys on either side so $4 choose 2$
And finally the other 3 can be arranged in $3! ways$
Pick two adjacent slots from your 7 $6choose 1$
then put the two girls in and multiply by 2 as they can be arranged in 2 ways
Then select the adjacent boys and again multiply by 2 and finally arrange the last.(account for rotations) This gives;
$$frac{{6choose 1}cdot {3choose 2}cdot 2 cdot {4choose 2}cdot 2 cdot 3!}{6}$$
ways and hence the probability is the second expression divided by the first.
answered Dec 6 '15 at 2:44
KingJ
511212
I think you should review your solution. The answer of $0.3$ it yields doesn't seem correct.
– true blue anil
Dec 6 '15 at 6:49
Ah yes, forgot to include the second 2 should be fixed now
– KingJ
Dec 6 '15 at 18:29
add a comment |
a bit off topic but are you applying to Oxbridge?
And now to the question;
The total # ways of arranging is$frac{7!}{7}$ (due to rotations)
The # of ways of putting two girls together is $3choose 2$
Now they must have boys on either side so $4 choose 2$
And finally the other 3 can be arranged in $3! ways$
Pick two adjacent slots from your 7 $6choose 1$
then put the two girls in and multiply by 2 as they can be arranged in 2 ways
Then select the adjacent boys and again multiply by 2 and finally arrange the last.(account for rotations) This gives;
$$frac{{6choose 1}cdot {3choose 2}cdot 2 cdot {4choose 2}cdot 2 cdot 3!}{6}$$
ways and hence the probability is the second expression divided by the first.
answered Dec 6 '15 at 2:44
KingJ
511212
I think you should review your solution. The answer of $0.3$ it yields doesn't seem correct.
– true blue anil
Dec 6 '15 at 6:49
Ah yes, forgot to include the second 2 should be fixed now
– KingJ
Dec 6 '15 at 18:29
add a comment |
a bit off topic but are you applying to Oxbridge?
And now to the question;
The total # ways of arranging is$frac{7!}{7}$ (due to rotations)
The # of ways of putting two girls together is $3choose 2$
Now they must have boys on either side so $4 choose 2$
And finally the other 3 can be arranged in $3! ways$
Pick two adjacent slots from your 7 $6choose 1$
then put the two girls in and multiply by 2 as they can be arranged in 2 ways
Then select the adjacent boys and again multiply by 2 and finally arrange the last.(account for rotations) This gives;
$$frac{{6choose 1}cdot {3choose 2}cdot 2 cdot {4choose 2}cdot 2 cdot 3!}{6}$$
ways and hence the probability is the second expression divided by the first.
answered Dec 6 '15 at 2:44
KingJ
511212
a bit off topic but are you applying to Oxbridge?
And now to the question;
The total # ways of arranging is$frac{7!}{7}$ (due to rotations)
The # of ways of putting two girls together is $3choose 2$
Now they must have boys on either side so $4 choose 2$
And finally the other 3 can be arranged in $3! ways$
Pick two adjacent slots from your 7 $6choose 1$
then put the two girls in and multiply by 2 as they can be arranged in 2 ways
Then select the adjacent boys and again multiply by 2 and finally arrange the last.(account for rotations) This gives;
$$frac{{6choose 1}cdot {3choose 2}cdot 2 cdot {4choose 2}cdot 2 cdot 3!}{6}$$
ways and hence the probability is the second expression divided by the first.
answered Dec 6 '15 at 2:44
KingJ
511212
edited Dec 6 '15 at 18:29
answered Dec 6 '15 at 2:44
KingJ
511212
answered Dec 6 '15 at 2:44
KingJ
511212
answered Dec 6 '15 at 2:44
KingJ
511212
511212
I think you should review your solution. The answer of $0.3$ it yields doesn't seem correct.
– true blue anil
Dec 6 '15 at 6:49
Ah yes, forgot to include the second 2 should be fixed now
– KingJ
Dec 6 '15 at 18:29
add a comment |
I think you should review your solution. The answer of $0.3$ it yields doesn't seem correct.
– true blue anil
Dec 6 '15 at 6:49
Ah yes, forgot to include the second 2 should be fixed now
– KingJ
Dec 6 '15 at 18:29
I think you should review your solution. The answer of $0.3$ it yields doesn't seem correct.
– true blue anil
Dec 6 '15 at 6:49
I think you should review your solution. The answer of $0.3$ it yields doesn't seem correct.
– true blue anil
Dec 6 '15 at 6:49
Ah yes, forgot to include the second 2 should be fixed now
– KingJ
Dec 6 '15 at 18:29
Ah yes, forgot to include the second 2 should be fixed now
– KingJ
Dec 6 '15 at 18:29
add a comment |
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