Mixing
![Object member as a parameter of a function in C# [duplicate]](https://lh3.googleusercontent.com/-tTw3-XCa-dc/AAAAAAAAAAI/AAAAAAAAAAA/AGDgw-goDtG5mxH4U1H5LDpXkt1qbaPt_A/s72-c-mo/photo.jpg?sz=32)
Two ends are occupied by boys and no girls are adjacent.
$begingroup$
I am learning combinatorics from the book Principles and Techniques in Combinatorics.
I am having some trouble understanding why my solution to the following problem is incorrect:
Example 1.2.3. There are 7 boys and 3 girls in a gathering. In how
many ways can they be arranged in a row so that
(i) the 3 girls form a single block (i.e. there is no boy between any two of
the girls)?
(ii) the two end-positions are occupied by boys and no girls are adjacent?
The second part is what I am having trouble with.
I first arrange two boys in the two extreme positions. There are $P(7,2)$ ways of doing this. Then for the remaining eight positions, I calculate the number of permutations of the boys and girls in which the girls are not adjacent to each other like so $8!-6!times3!$.
Then by the multiplication principle, we have $P(7,2)times(8!-6!times3!)$ such arrangements.
Please note that I do understand the solution where the boys are arranged first and the girls are placed between them afterwards. I just want to know what is wrong with mine and what I can do prevent such a mistake from happening again.
combinatorics
edited Jan 18 at 17:05
N. F. Taussig
44.5k103357
asked Jan 18 at 14:47
DS2830DS2830
183
$endgroup$
|
show 4 more comments
$begingroup$
I am learning combinatorics from the book Principles and Techniques in Combinatorics.
I am having some trouble understanding why my solution to the following problem is incorrect:
Example 1.2.3. There are 7 boys and 3 girls in a gathering. In how
many ways can they be arranged in a row so that
(i) the 3 girls form a single block (i.e. there is no boy between any two of
the girls)?
(ii) the two end-positions are occupied by boys and no girls are adjacent?
The second part is what I am having trouble with.
I first arrange two boys in the two extreme positions. There are $P(7,2)$ ways of doing this. Then for the remaining eight positions, I calculate the number of permutations of the boys and girls in which the girls are not adjacent to each other like so $8!-6!times3!$.
Then by the multiplication principle, we have $P(7,2)times(8!-6!times3!)$ such arrangements.
Please note that I do understand the solution where the boys are arranged first and the girls are placed between them afterwards. I just want to know what is wrong with mine and what I can do prevent such a mistake from happening again.
combinatorics
edited Jan 18 at 17:05
N. F. Taussig
44.5k103357
asked Jan 18 at 14:47
DS2830DS2830
183
$endgroup$
$begingroup$
Your expression just counts the ways to do it so that the three girls aren't all together in a block. You could still have, e.g. $GGBGcdots$.
$endgroup$
– lulu
Jan 18 at 15:00
$begingroup$
I see. Would using the inclusion-exclusion principle to calculate the permutations in which neither 2 nor 3 girls are together and then subtracting that from $8!$ work?
$endgroup$
– DS2830
Jan 18 at 15:09
$begingroup$
I prefer the stars and bars method. You have gaps around and between the girls, as $_G_G_G_$. You need to put at least one boy in each of the two middle gaps, but then you have $3$ boys left to distribute to $4$ gaps without restriction.
$endgroup$
– lulu
Jan 18 at 15:11
$begingroup$
I see. I don't have much experience with the stars and bars method. However, I do know some inclusion-exclusion, would that be applicable here? If yes, I'm thinking of finding $|AUB|$ where $A,B$ represent the events of two and three girls being together respectively and then subtracting that from $8!$.
$endgroup$
– DS2830
Jan 18 at 15:19
$begingroup$
Yes, you could do it by inclusion exclusion. First subtract off the orders in which any specified pair of girls sit together. Then add back the orders in which all three sit together.
$endgroup$
– lulu
Jan 18 at 15:20
|
show 4 more comments
$begingroup$
I am learning combinatorics from the book Principles and Techniques in Combinatorics.
I am having some trouble understanding why my solution to the following problem is incorrect:
Example 1.2.3. There are 7 boys and 3 girls in a gathering. In how
many ways can they be arranged in a row so that
(i) the 3 girls form a single block (i.e. there is no boy between any two of
the girls)?
(ii) the two end-positions are occupied by boys and no girls are adjacent?
The second part is what I am having trouble with.
I first arrange two boys in the two extreme positions. There are $P(7,2)$ ways of doing this. Then for the remaining eight positions, I calculate the number of permutations of the boys and girls in which the girls are not adjacent to each other like so $8!-6!times3!$.
Then by the multiplication principle, we have $P(7,2)times(8!-6!times3!)$ such arrangements.
Please note that I do understand the solution where the boys are arranged first and the girls are placed between them afterwards. I just want to know what is wrong with mine and what I can do prevent such a mistake from happening again.
combinatorics
edited Jan 18 at 17:05
N. F. Taussig
44.5k103357
asked Jan 18 at 14:47
DS2830DS2830
183
$endgroup$
I am learning combinatorics from the book Principles and Techniques in Combinatorics.
I am having some trouble understanding why my solution to the following problem is incorrect:
Example 1.2.3. There are 7 boys and 3 girls in a gathering. In how
many ways can they be arranged in a row so that
(i) the 3 girls form a single block (i.e. there is no boy between any two of
the girls)?
(ii) the two end-positions are occupied by boys and no girls are adjacent?
The second part is what I am having trouble with.
I first arrange two boys in the two extreme positions. There are $P(7,2)$ ways of doing this. Then for the remaining eight positions, I calculate the number of permutations of the boys and girls in which the girls are not adjacent to each other like so $8!-6!times3!$.
Then by the multiplication principle, we have $P(7,2)times(8!-6!times3!)$ such arrangements.
Please note that I do understand the solution where the boys are arranged first and the girls are placed between them afterwards. I just want to know what is wrong with mine and what I can do prevent such a mistake from happening again.
combinatorics
combinatorics
edited Jan 18 at 17:05
N. F. Taussig
44.5k103357
asked Jan 18 at 14:47
DS2830DS2830
183
edited Jan 18 at 17:05
N. F. Taussig
44.5k103357
asked Jan 18 at 14:47
DS2830DS2830
183
edited Jan 18 at 17:05
N. F. Taussig
44.5k103357
edited Jan 18 at 17:05
N. F. Taussig
44.5k103357
edited Jan 18 at 17:05
N. F. Taussig
44.5k103357
44.5k103357
asked Jan 18 at 14:47
DS2830DS2830
183
asked Jan 18 at 14:47
DS2830DS2830
183
asked Jan 18 at 14:47
DS2830DS2830
183
183
$begingroup$
Your expression just counts the ways to do it so that the three girls aren't all together in a block. You could still have, e.g. $GGBGcdots$.
$endgroup$
– lulu
Jan 18 at 15:00
$begingroup$
I see. Would using the inclusion-exclusion principle to calculate the permutations in which neither 2 nor 3 girls are together and then subtracting that from $8!$ work?
$endgroup$
– DS2830
Jan 18 at 15:09
$begingroup$
I prefer the stars and bars method. You have gaps around and between the girls, as $_G_G_G_$. You need to put at least one boy in each of the two middle gaps, but then you have $3$ boys left to distribute to $4$ gaps without restriction.
$endgroup$
– lulu
Jan 18 at 15:11
$begingroup$
I see. I don't have much experience with the stars and bars method. However, I do know some inclusion-exclusion, would that be applicable here? If yes, I'm thinking of finding $|AUB|$ where $A,B$ represent the events of two and three girls being together respectively and then subtracting that from $8!$.
$endgroup$
– DS2830
Jan 18 at 15:19
$begingroup$
Yes, you could do it by inclusion exclusion. First subtract off the orders in which any specified pair of girls sit together. Then add back the orders in which all three sit together.
$endgroup$
– lulu
Jan 18 at 15:20
|
show 4 more comments
$begingroup$
Your expression just counts the ways to do it so that the three girls aren't all together in a block. You could still have, e.g. $GGBGcdots$.
$endgroup$
– lulu
Jan 18 at 15:00
$begingroup$
I see. Would using the inclusion-exclusion principle to calculate the permutations in which neither 2 nor 3 girls are together and then subtracting that from $8!$ work?
$endgroup$
– DS2830
Jan 18 at 15:09
$begingroup$
I prefer the stars and bars method. You have gaps around and between the girls, as $_G_G_G_$. You need to put at least one boy in each of the two middle gaps, but then you have $3$ boys left to distribute to $4$ gaps without restriction.
$endgroup$
– lulu
Jan 18 at 15:11
$begingroup$
I see. I don't have much experience with the stars and bars method. However, I do know some inclusion-exclusion, would that be applicable here? If yes, I'm thinking of finding $|AUB|$ where $A,B$ represent the events of two and three girls being together respectively and then subtracting that from $8!$.
$endgroup$
– DS2830
Jan 18 at 15:19
$begingroup$
Yes, you could do it by inclusion exclusion. First subtract off the orders in which any specified pair of girls sit together. Then add back the orders in which all three sit together.
$endgroup$
– lulu
Jan 18 at 15:20
$begingroup$
Your expression just counts the ways to do it so that the three girls aren't all together in a block. You could still have, e.g. $GGBGcdots$.
$endgroup$
– lulu
Jan 18 at 15:00
$begingroup$
Your expression just counts the ways to do it so that the three girls aren't all together in a block. You could still have, e.g. $GGBGcdots$.
$endgroup$
– lulu
Jan 18 at 15:00
$begingroup$
I see. Would using the inclusion-exclusion principle to calculate the permutations in which neither 2 nor 3 girls are together and then subtracting that from $8!$ work?
$endgroup$
– DS2830
Jan 18 at 15:09
$begingroup$
I see. Would using the inclusion-exclusion principle to calculate the permutations in which neither 2 nor 3 girls are together and then subtracting that from $8!$ work?
$endgroup$
– DS2830
Jan 18 at 15:09
$begingroup$
I prefer the stars and bars method. You have gaps around and between the girls, as $_G_G_G_$. You need to put at least one boy in each of the two middle gaps, but then you have $3$ boys left to distribute to $4$ gaps without restriction.
$endgroup$
– lulu
Jan 18 at 15:11
$begingroup$
I prefer the stars and bars method. You have gaps around and between the girls, as $_G_G_G_$. You need to put at least one boy in each of the two middle gaps, but then you have $3$ boys left to distribute to $4$ gaps without restriction.
$endgroup$
– lulu
Jan 18 at 15:11
$begingroup$
I see. I don't have much experience with the stars and bars method. However, I do know some inclusion-exclusion, would that be applicable here? If yes, I'm thinking of finding $|AUB|$ where $A,B$ represent the events of two and three girls being together respectively and then subtracting that from $8!$.
$endgroup$
– DS2830
Jan 18 at 15:19
$begingroup$
I see. I don't have much experience with the stars and bars method. However, I do know some inclusion-exclusion, would that be applicable here? If yes, I'm thinking of finding $|AUB|$ where $A,B$ represent the events of two and three girls being together respectively and then subtracting that from $8!$.
$endgroup$
– DS2830
Jan 18 at 15:19
$begingroup$
Yes, you could do it by inclusion exclusion. First subtract off the orders in which any specified pair of girls sit together. Then add back the orders in which all three sit together.
$endgroup$
– lulu
Jan 18 at 15:20
$begingroup$
Yes, you could do it by inclusion exclusion. First subtract off the orders in which any specified pair of girls sit together. Then add back the orders in which all three sit together.
$endgroup$
– lulu
Jan 18 at 15:20
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Arrange the seven boys in a row, which can be done in $7!$ ways. Since boys must occupy both ends of the row, this creates six spaces in which the three girls can be placed.
$$B square B square B square B square B square B square B$$
To ensure that no two girls are adjacent, choose three of those six spaces in which to place a girl, which can be done in $binom{6}{3}$ ways. The three girls can be arranged in the selected spaces in $3!$ ways. Hence, the number of admissible arrangements is
$$7!binom{6}{3}3!$$
answered Jan 18 at 17:09
N. F. TaussigN. F. Taussig
44.5k103357
$endgroup$
add a comment |
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$begingroup$
Arrange the seven boys in a row, which can be done in $7!$ ways. Since boys must occupy both ends of the row, this creates six spaces in which the three girls can be placed.
$$B square B square B square B square B square B square B$$
To ensure that no two girls are adjacent, choose three of those six spaces in which to place a girl, which can be done in $binom{6}{3}$ ways. The three girls can be arranged in the selected spaces in $3!$ ways. Hence, the number of admissible arrangements is
$$7!binom{6}{3}3!$$
answered Jan 18 at 17:09
N. F. TaussigN. F. Taussig
44.5k103357
$endgroup$
add a comment |
$begingroup$
Arrange the seven boys in a row, which can be done in $7!$ ways. Since boys must occupy both ends of the row, this creates six spaces in which the three girls can be placed.
$$B square B square B square B square B square B square B$$
To ensure that no two girls are adjacent, choose three of those six spaces in which to place a girl, which can be done in $binom{6}{3}$ ways. The three girls can be arranged in the selected spaces in $3!$ ways. Hence, the number of admissible arrangements is
$$7!binom{6}{3}3!$$
answered Jan 18 at 17:09
N. F. TaussigN. F. Taussig
44.5k103357
$endgroup$
add a comment |
$begingroup$
Arrange the seven boys in a row, which can be done in $7!$ ways. Since boys must occupy both ends of the row, this creates six spaces in which the three girls can be placed.
$$B square B square B square B square B square B square B$$
To ensure that no two girls are adjacent, choose three of those six spaces in which to place a girl, which can be done in $binom{6}{3}$ ways. The three girls can be arranged in the selected spaces in $3!$ ways. Hence, the number of admissible arrangements is
$$7!binom{6}{3}3!$$
answered Jan 18 at 17:09
N. F. TaussigN. F. Taussig
44.5k103357
$endgroup$
Arrange the seven boys in a row, which can be done in $7!$ ways. Since boys must occupy both ends of the row, this creates six spaces in which the three girls can be placed.
$$B square B square B square B square B square B square B$$
To ensure that no two girls are adjacent, choose three of those six spaces in which to place a girl, which can be done in $binom{6}{3}$ ways. The three girls can be arranged in the selected spaces in $3!$ ways. Hence, the number of admissible arrangements is
$$7!binom{6}{3}3!$$
answered Jan 18 at 17:09
N. F. TaussigN. F. Taussig
44.5k103357
answered Jan 18 at 17:09
N. F. TaussigN. F. Taussig
44.5k103357
answered Jan 18 at 17:09
N. F. TaussigN. F. Taussig
44.5k103357
answered Jan 18 at 17:09
N. F. TaussigN. F. Taussig
44.5k103357
44.5k103357
add a comment |
add a comment |
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$begingroup$
Your expression just counts the ways to do it so that the three girls aren't all together in a block. You could still have, e.g. $GGBGcdots$.
$endgroup$
– lulu
Jan 18 at 15:00
$begingroup$
I see. Would using the inclusion-exclusion principle to calculate the permutations in which neither 2 nor 3 girls are together and then subtracting that from $8!$ work?
$endgroup$
– DS2830
Jan 18 at 15:09
$begingroup$
I prefer the stars and bars method. You have gaps around and between the girls, as $_G_G_G_$. You need to put at least one boy in each of the two middle gaps, but then you have $3$ boys left to distribute to $4$ gaps without restriction.
$endgroup$
– lulu
Jan 18 at 15:11
$begingroup$
I see. I don't have much experience with the stars and bars method. However, I do know some inclusion-exclusion, would that be applicable here? If yes, I'm thinking of finding $|AUB|$ where $A,B$ represent the events of two and three girls being together respectively and then subtracting that from $8!$.
$endgroup$
– DS2830
Jan 18 at 15:19
$begingroup$
Yes, you could do it by inclusion exclusion. First subtract off the orders in which any specified pair of girls sit together. Then add back the orders in which all three sit together.
$endgroup$
– lulu
Jan 18 at 15:20