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4 females and 6 males will be seated on 19 chairs
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There are 4 females and 6 males students. They will be seated on 19 chairs. How many ways can we do this, if no two females are seated in adjacent?
What makes me confuse is there are only 10 students but the chairs are 19.
If the chairs are 10 only (equal to the number of persons) I could do C(7,4)*4!6!
combinatorics permutations
asked 2 hours ago
Muhamad Abdul Rosid
475
add a comment |Â
up vote
5
down vote
favorite
There are 4 females and 6 males students. They will be seated on 19 chairs. How many ways can we do this, if no two females are seated in adjacent?
What makes me confuse is there are only 10 students but the chairs are 19.
If the chairs are 10 only (equal to the number of persons) I could do C(7,4)*4!6!
combinatorics permutations
asked 2 hours ago
Muhamad Abdul Rosid
475
So there are a lot of empty seats. For the first part, just place the women, so $binom 194$, and then place the men, $binom 156$. (and then assign people). Can you finish from here?
– lulu
2 hours ago
But, the females are not adjacent.
– Muhamad Abdul Rosid
50 mins ago
Do you have the answer? Compare it with 157405248000, if it's so, trust my answer.
– Isana Yashiro
40 mins ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
There are 4 females and 6 males students. They will be seated on 19 chairs. How many ways can we do this, if no two females are seated in adjacent?
What makes me confuse is there are only 10 students but the chairs are 19.
If the chairs are 10 only (equal to the number of persons) I could do C(7,4)*4!6!
combinatorics permutations
asked 2 hours ago
Muhamad Abdul Rosid
475
There are 4 females and 6 males students. They will be seated on 19 chairs. How many ways can we do this, if no two females are seated in adjacent?
What makes me confuse is there are only 10 students but the chairs are 19.
If the chairs are 10 only (equal to the number of persons) I could do C(7,4)*4!6!
combinatorics permutations
combinatorics permutations
asked 2 hours ago
Muhamad Abdul Rosid
475
asked 2 hours ago
Muhamad Abdul Rosid
475
asked 2 hours ago
Muhamad Abdul Rosid
475
asked 2 hours ago
Muhamad Abdul Rosid
475
asked 2 hours ago
Muhamad Abdul Rosid
475
475
So there are a lot of empty seats. For the first part, just place the women, so $binom 194$, and then place the men, $binom 156$. (and then assign people). Can you finish from here?
– lulu
2 hours ago
But, the females are not adjacent.
– Muhamad Abdul Rosid
50 mins ago
Do you have the answer? Compare it with 157405248000, if it's so, trust my answer.
– Isana Yashiro
40 mins ago
add a comment |Â
So there are a lot of empty seats. For the first part, just place the women, so $binom 194$, and then place the men, $binom 156$. (and then assign people). Can you finish from here?
– lulu
2 hours ago
But, the females are not adjacent.
– Muhamad Abdul Rosid
50 mins ago
Do you have the answer? Compare it with 157405248000, if it's so, trust my answer.
– Isana Yashiro
40 mins ago
So there are a lot of empty seats. For the first part, just place the women, so $binom 194$, and then place the men, $binom 156$. (and then assign people). Can you finish from here?
– lulu
2 hours ago
So there are a lot of empty seats. For the first part, just place the women, so $binom 194$, and then place the men, $binom 156$. (and then assign people). Can you finish from here?
– lulu
2 hours ago
But, the females are not adjacent.
– Muhamad Abdul Rosid
50 mins ago
But, the females are not adjacent.
– Muhamad Abdul Rosid
50 mins ago
Do you have the answer? Compare it with 157405248000, if it's so, trust my answer.
– Isana Yashiro
40 mins ago
Do you have the answer? Compare it with 157405248000, if it's so, trust my answer.
– Isana Yashiro
40 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
Total Number of chairs=19.
Total number of persons=10.
So number of possible to select 10 chairs from 19 chairs $=19C10$.
Now take 10 chairs.
Then
$_B_B_B_B_B_B_$
If you're filling 4 girls in any 4 spaces among 7 spaces yiels a required arrangements.
There are $7C4$ ways to choose filling 4 girls.
Thus totally we have
$19C10×7C4×4!×6!$ ways.
Where $4!$ represents PERMUTATION among 4 girls and $6!$ represents PERMUTATION among 6 boys.
answered 37 mins ago
Avinash N
712411
I think you miss a case: if [girl, empty, girl], then there are no adjacent girls. That is: if the choice of those 10 chairs in you $19C10$, accidentally, choose some two chairs that are not adjacent, then even if your $7C4$ have two girl (seems like) adjacent, they actually not.
– Isana Yashiro
31 mins ago
@IsanaYashiro if the empty girl means you leave the space and merge the boys
– Avinash N
28 mins ago
sorry, it's a typo, I meant girl - empty - girl.
– Isana Yashiro
27 mins ago
1
@IsanaYashiro yes sir. I got it. Thank you so much. I will fix this issue soon. Once again thank you.
– Avinash N
5 mins ago
1
No worries, cheers for the love for combinatorics~
– Isana Yashiro
3 mins ago
 |Â
show 2 more comments
up vote
2
down vote
Let's first sit the females... For that first select $4$ chairs to them: $19 choose 4$. Now we have left $19-4$ chairs remaining. Let's then distribute the remaining chairs between the females to help us with the restriction that no two females are seated in adjacent... For that, think as the females as separators, therefore:
$$
underbrace _textslot 1text female 1underbrace _textslot 2text female 2underbrace _textslot 3text female 3underbrace _textslot 4text female 4underbrace _textslot 5
$$
where slots 2, 3 and 4 need to have at least one chair. From that we can create the following equation:
$$
underbraces_1_geq 0 +underbraces_2_geq 1 +underbraces_3_geq 1+underbraces_4_geq 1+
underbraces_5_geq 0 = 19-4 = 15
$$
using stars and bars method we get that we can distribute the chairs in those slots in $16 choose 4$ ways.
Now we've just sat 4 females and distributed the remaining $15$ chairs according to our restriction. Let's then choose $6$ of those $15$ chairs to sit the males. We can do that in $15 choose 6$ ways.
Finally we have males and females sitting in a way that between two females we're going to have or chairs or males to separate them. We just need to organize them in all possible ways now: $4!6!$.
And our final answer, by multiplication principle is:
$$
19 choose 416 choose 415 choose 64!6!
$$
answered 28 mins ago
Bruno Reis
854417
I think you made a mistake: if your $19choose4$ choose position(s) at the left end and/or the right end, then your assumption of slot1 slot5 not exists
– Isana Yashiro
19 mins ago
@Isana Yashiro No, they do exists, but they don't have the positivity restriction.
– Bruno Reis
6 mins ago
add a comment |Â
up vote
2
down vote
Here is one way of thinking it: First seat the men then seat the women. And arrange people in the right order before seating.
beginalignP(15,6)cdot P(16,4)\=boxed15choose66!cdot16choose44!endalign
It may sounds creepy, but here is another hint if you don't get it:
beginaligntextrmBeyond those people, there are identical 9 ghost there(to help you)endalign
The key idea is about the $15$ and $16$:
- 15: this comes from 19-4, why? because I first consider those 6 men and 9 ghosts.
- 16: now we finish the queuing of 6+9=15 men-ghost-queue, what's the next step to make girls no adjacent?
answered 51 mins ago
Isana Yashiro
1,2801622
But the females are not adjacent
– Muhamad Abdul Rosid
49 mins ago
@MuhamadAbdulRosid: That's considered in my answer.
– Isana Yashiro
44 mins ago
I know there are 9 ghosts. But, why is it C(15,6) why 15? I don't understand. Then why C(16,4)? Where does it come from? The 16
– Muhamad Abdul Rosid
11 mins ago
@MuhamadAbdulRosid: Correct, that's what want you to think, that's the key idea/value of this problem, you're on the right path, I will edit for clarification.
– Isana Yashiro
9 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Total Number of chairs=19.
Total number of persons=10.
So number of possible to select 10 chairs from 19 chairs $=19C10$.
Now take 10 chairs.
Then
$_B_B_B_B_B_B_$
If you're filling 4 girls in any 4 spaces among 7 spaces yiels a required arrangements.
There are $7C4$ ways to choose filling 4 girls.
Thus totally we have
$19C10×7C4×4!×6!$ ways.
Where $4!$ represents PERMUTATION among 4 girls and $6!$ represents PERMUTATION among 6 boys.
answered 37 mins ago
Avinash N
712411
I think you miss a case: if [girl, empty, girl], then there are no adjacent girls. That is: if the choice of those 10 chairs in you $19C10$, accidentally, choose some two chairs that are not adjacent, then even if your $7C4$ have two girl (seems like) adjacent, they actually not.
– Isana Yashiro
31 mins ago
@IsanaYashiro if the empty girl means you leave the space and merge the boys
– Avinash N
28 mins ago
sorry, it's a typo, I meant girl - empty - girl.
– Isana Yashiro
27 mins ago
1
@IsanaYashiro yes sir. I got it. Thank you so much. I will fix this issue soon. Once again thank you.
– Avinash N
5 mins ago
1
No worries, cheers for the love for combinatorics~
– Isana Yashiro
3 mins ago
 |Â
show 2 more comments
up vote
2
down vote
Total Number of chairs=19.
Total number of persons=10.
So number of possible to select 10 chairs from 19 chairs $=19C10$.
Now take 10 chairs.
Then
$_B_B_B_B_B_B_$
If you're filling 4 girls in any 4 spaces among 7 spaces yiels a required arrangements.
There are $7C4$ ways to choose filling 4 girls.
Thus totally we have
$19C10×7C4×4!×6!$ ways.
Where $4!$ represents PERMUTATION among 4 girls and $6!$ represents PERMUTATION among 6 boys.
answered 37 mins ago
Avinash N
712411
I think you miss a case: if [girl, empty, girl], then there are no adjacent girls. That is: if the choice of those 10 chairs in you $19C10$, accidentally, choose some two chairs that are not adjacent, then even if your $7C4$ have two girl (seems like) adjacent, they actually not.
– Isana Yashiro
31 mins ago
@IsanaYashiro if the empty girl means you leave the space and merge the boys
– Avinash N
28 mins ago
sorry, it's a typo, I meant girl - empty - girl.
– Isana Yashiro
27 mins ago
1
@IsanaYashiro yes sir. I got it. Thank you so much. I will fix this issue soon. Once again thank you.
– Avinash N
5 mins ago
1
No worries, cheers for the love for combinatorics~
– Isana Yashiro
3 mins ago
 |Â
show 2 more comments
up vote
2
down vote
up vote
2
down vote
Total Number of chairs=19.
Total number of persons=10.
So number of possible to select 10 chairs from 19 chairs $=19C10$.
Now take 10 chairs.
Then
$_B_B_B_B_B_B_$
If you're filling 4 girls in any 4 spaces among 7 spaces yiels a required arrangements.
There are $7C4$ ways to choose filling 4 girls.
Thus totally we have
$19C10×7C4×4!×6!$ ways.
Where $4!$ represents PERMUTATION among 4 girls and $6!$ represents PERMUTATION among 6 boys.
answered 37 mins ago
Avinash N
712411
Total Number of chairs=19.
Total number of persons=10.
So number of possible to select 10 chairs from 19 chairs $=19C10$.
Now take 10 chairs.
Then
$_B_B_B_B_B_B_$
If you're filling 4 girls in any 4 spaces among 7 spaces yiels a required arrangements.
There are $7C4$ ways to choose filling 4 girls.
Thus totally we have
$19C10×7C4×4!×6!$ ways.
Where $4!$ represents PERMUTATION among 4 girls and $6!$ represents PERMUTATION among 6 boys.
answered 37 mins ago
Avinash N
712411
answered 37 mins ago
Avinash N
712411
answered 37 mins ago
Avinash N
712411
answered 37 mins ago
Avinash N
712411
712411
I think you miss a case: if [girl, empty, girl], then there are no adjacent girls. That is: if the choice of those 10 chairs in you $19C10$, accidentally, choose some two chairs that are not adjacent, then even if your $7C4$ have two girl (seems like) adjacent, they actually not.
– Isana Yashiro
31 mins ago
@IsanaYashiro if the empty girl means you leave the space and merge the boys
– Avinash N
28 mins ago
sorry, it's a typo, I meant girl - empty - girl.
– Isana Yashiro
27 mins ago
1
@IsanaYashiro yes sir. I got it. Thank you so much. I will fix this issue soon. Once again thank you.
– Avinash N
5 mins ago
1
No worries, cheers for the love for combinatorics~
– Isana Yashiro
3 mins ago
 |Â
show 2 more comments
I think you miss a case: if [girl, empty, girl], then there are no adjacent girls. That is: if the choice of those 10 chairs in you $19C10$, accidentally, choose some two chairs that are not adjacent, then even if your $7C4$ have two girl (seems like) adjacent, they actually not.
– Isana Yashiro
31 mins ago
@IsanaYashiro if the empty girl means you leave the space and merge the boys
– Avinash N
28 mins ago
sorry, it's a typo, I meant girl - empty - girl.
– Isana Yashiro
27 mins ago
1
@IsanaYashiro yes sir. I got it. Thank you so much. I will fix this issue soon. Once again thank you.
– Avinash N
5 mins ago
1
No worries, cheers for the love for combinatorics~
– Isana Yashiro
3 mins ago
I think you miss a case: if [girl, empty, girl], then there are no adjacent girls. That is: if the choice of those 10 chairs in you $19C10$, accidentally, choose some two chairs that are not adjacent, then even if your $7C4$ have two girl (seems like) adjacent, they actually not.
– Isana Yashiro
31 mins ago
I think you miss a case: if [girl, empty, girl], then there are no adjacent girls. That is: if the choice of those 10 chairs in you $19C10$, accidentally, choose some two chairs that are not adjacent, then even if your $7C4$ have two girl (seems like) adjacent, they actually not.
– Isana Yashiro
31 mins ago
@IsanaYashiro if the empty girl means you leave the space and merge the boys
– Avinash N
28 mins ago
@IsanaYashiro if the empty girl means you leave the space and merge the boys
– Avinash N
28 mins ago
sorry, it's a typo, I meant girl - empty - girl.
– Isana Yashiro
27 mins ago
sorry, it's a typo, I meant girl - empty - girl.
– Isana Yashiro
27 mins ago
1
1
@IsanaYashiro yes sir. I got it. Thank you so much. I will fix this issue soon. Once again thank you.
– Avinash N
5 mins ago
@IsanaYashiro yes sir. I got it. Thank you so much. I will fix this issue soon. Once again thank you.
– Avinash N
5 mins ago
1
1
No worries, cheers for the love for combinatorics~
– Isana Yashiro
3 mins ago
No worries, cheers for the love for combinatorics~
– Isana Yashiro
3 mins ago
 |Â
show 2 more comments
up vote
2
down vote
Let's first sit the females... For that first select $4$ chairs to them: $19 choose 4$. Now we have left $19-4$ chairs remaining. Let's then distribute the remaining chairs between the females to help us with the restriction that no two females are seated in adjacent... For that, think as the females as separators, therefore:
$$
underbrace _textslot 1text female 1underbrace _textslot 2text female 2underbrace _textslot 3text female 3underbrace _textslot 4text female 4underbrace _textslot 5
$$
where slots 2, 3 and 4 need to have at least one chair. From that we can create the following equation:
$$
underbraces_1_geq 0 +underbraces_2_geq 1 +underbraces_3_geq 1+underbraces_4_geq 1+
underbraces_5_geq 0 = 19-4 = 15
$$
using stars and bars method we get that we can distribute the chairs in those slots in $16 choose 4$ ways.
Now we've just sat 4 females and distributed the remaining $15$ chairs according to our restriction. Let's then choose $6$ of those $15$ chairs to sit the males. We can do that in $15 choose 6$ ways.
Finally we have males and females sitting in a way that between two females we're going to have or chairs or males to separate them. We just need to organize them in all possible ways now: $4!6!$.
And our final answer, by multiplication principle is:
$$
19 choose 416 choose 415 choose 64!6!
$$
answered 28 mins ago
Bruno Reis
854417
I think you made a mistake: if your $19choose4$ choose position(s) at the left end and/or the right end, then your assumption of slot1 slot5 not exists
– Isana Yashiro
19 mins ago
@Isana Yashiro No, they do exists, but they don't have the positivity restriction.
– Bruno Reis
6 mins ago
add a comment |Â
up vote
2
down vote
Let's first sit the females... For that first select $4$ chairs to them: $19 choose 4$. Now we have left $19-4$ chairs remaining. Let's then distribute the remaining chairs between the females to help us with the restriction that no two females are seated in adjacent... For that, think as the females as separators, therefore:
$$
underbrace _textslot 1text female 1underbrace _textslot 2text female 2underbrace _textslot 3text female 3underbrace _textslot 4text female 4underbrace _textslot 5
$$
where slots 2, 3 and 4 need to have at least one chair. From that we can create the following equation:
$$
underbraces_1_geq 0 +underbraces_2_geq 1 +underbraces_3_geq 1+underbraces_4_geq 1+
underbraces_5_geq 0 = 19-4 = 15
$$
using stars and bars method we get that we can distribute the chairs in those slots in $16 choose 4$ ways.
Now we've just sat 4 females and distributed the remaining $15$ chairs according to our restriction. Let's then choose $6$ of those $15$ chairs to sit the males. We can do that in $15 choose 6$ ways.
Finally we have males and females sitting in a way that between two females we're going to have or chairs or males to separate them. We just need to organize them in all possible ways now: $4!6!$.
And our final answer, by multiplication principle is:
$$
19 choose 416 choose 415 choose 64!6!
$$
answered 28 mins ago
Bruno Reis
854417
I think you made a mistake: if your $19choose4$ choose position(s) at the left end and/or the right end, then your assumption of slot1 slot5 not exists
– Isana Yashiro
19 mins ago
@Isana Yashiro No, they do exists, but they don't have the positivity restriction.
– Bruno Reis
6 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let's first sit the females... For that first select $4$ chairs to them: $19 choose 4$. Now we have left $19-4$ chairs remaining. Let's then distribute the remaining chairs between the females to help us with the restriction that no two females are seated in adjacent... For that, think as the females as separators, therefore:
$$
underbrace _textslot 1text female 1underbrace _textslot 2text female 2underbrace _textslot 3text female 3underbrace _textslot 4text female 4underbrace _textslot 5
$$
where slots 2, 3 and 4 need to have at least one chair. From that we can create the following equation:
$$
underbraces_1_geq 0 +underbraces_2_geq 1 +underbraces_3_geq 1+underbraces_4_geq 1+
underbraces_5_geq 0 = 19-4 = 15
$$
using stars and bars method we get that we can distribute the chairs in those slots in $16 choose 4$ ways.
Now we've just sat 4 females and distributed the remaining $15$ chairs according to our restriction. Let's then choose $6$ of those $15$ chairs to sit the males. We can do that in $15 choose 6$ ways.
Finally we have males and females sitting in a way that between two females we're going to have or chairs or males to separate them. We just need to organize them in all possible ways now: $4!6!$.
And our final answer, by multiplication principle is:
$$
19 choose 416 choose 415 choose 64!6!
$$
answered 28 mins ago
Bruno Reis
854417
Let's first sit the females... For that first select $4$ chairs to them: $19 choose 4$. Now we have left $19-4$ chairs remaining. Let's then distribute the remaining chairs between the females to help us with the restriction that no two females are seated in adjacent... For that, think as the females as separators, therefore:
$$
underbrace _textslot 1text female 1underbrace _textslot 2text female 2underbrace _textslot 3text female 3underbrace _textslot 4text female 4underbrace _textslot 5
$$
where slots 2, 3 and 4 need to have at least one chair. From that we can create the following equation:
$$
underbraces_1_geq 0 +underbraces_2_geq 1 +underbraces_3_geq 1+underbraces_4_geq 1+
underbraces_5_geq 0 = 19-4 = 15
$$
using stars and bars method we get that we can distribute the chairs in those slots in $16 choose 4$ ways.
Now we've just sat 4 females and distributed the remaining $15$ chairs according to our restriction. Let's then choose $6$ of those $15$ chairs to sit the males. We can do that in $15 choose 6$ ways.
Finally we have males and females sitting in a way that between two females we're going to have or chairs or males to separate them. We just need to organize them in all possible ways now: $4!6!$.
And our final answer, by multiplication principle is:
$$
19 choose 416 choose 415 choose 64!6!
$$
answered 28 mins ago
Bruno Reis
854417
edited 13 mins ago
answered 28 mins ago
Bruno Reis
854417
answered 28 mins ago
Bruno Reis
854417
answered 28 mins ago
Bruno Reis
854417
854417
I think you made a mistake: if your $19choose4$ choose position(s) at the left end and/or the right end, then your assumption of slot1 slot5 not exists
– Isana Yashiro
19 mins ago
@Isana Yashiro No, they do exists, but they don't have the positivity restriction.
– Bruno Reis
6 mins ago
add a comment |Â
I think you made a mistake: if your $19choose4$ choose position(s) at the left end and/or the right end, then your assumption of slot1 slot5 not exists
– Isana Yashiro
19 mins ago
@Isana Yashiro No, they do exists, but they don't have the positivity restriction.
– Bruno Reis
6 mins ago
I think you made a mistake: if your $19choose4$ choose position(s) at the left end and/or the right end, then your assumption of slot1 slot5 not exists
– Isana Yashiro
19 mins ago
I think you made a mistake: if your $19choose4$ choose position(s) at the left end and/or the right end, then your assumption of slot1 slot5 not exists
– Isana Yashiro
19 mins ago
@Isana Yashiro No, they do exists, but they don't have the positivity restriction.
– Bruno Reis
6 mins ago
@Isana Yashiro No, they do exists, but they don't have the positivity restriction.
– Bruno Reis
6 mins ago
add a comment |Â
up vote
2
down vote
Here is one way of thinking it: First seat the men then seat the women. And arrange people in the right order before seating.
beginalignP(15,6)cdot P(16,4)\=boxed15choose66!cdot16choose44!endalign
It may sounds creepy, but here is another hint if you don't get it:
beginaligntextrmBeyond those people, there are identical 9 ghost there(to help you)endalign
The key idea is about the $15$ and $16$:
- 15: this comes from 19-4, why? because I first consider those 6 men and 9 ghosts.
- 16: now we finish the queuing of 6+9=15 men-ghost-queue, what's the next step to make girls no adjacent?
answered 51 mins ago
Isana Yashiro
1,2801622
But the females are not adjacent
– Muhamad Abdul Rosid
49 mins ago
@MuhamadAbdulRosid: That's considered in my answer.
– Isana Yashiro
44 mins ago
I know there are 9 ghosts. But, why is it C(15,6) why 15? I don't understand. Then why C(16,4)? Where does it come from? The 16
– Muhamad Abdul Rosid
11 mins ago
@MuhamadAbdulRosid: Correct, that's what want you to think, that's the key idea/value of this problem, you're on the right path, I will edit for clarification.
– Isana Yashiro
9 mins ago
add a comment |Â
up vote
2
down vote
Here is one way of thinking it: First seat the men then seat the women. And arrange people in the right order before seating.
beginalignP(15,6)cdot P(16,4)\=boxed15choose66!cdot16choose44!endalign
It may sounds creepy, but here is another hint if you don't get it:
beginaligntextrmBeyond those people, there are identical 9 ghost there(to help you)endalign
The key idea is about the $15$ and $16$:
- 15: this comes from 19-4, why? because I first consider those 6 men and 9 ghosts.
- 16: now we finish the queuing of 6+9=15 men-ghost-queue, what's the next step to make girls no adjacent?
answered 51 mins ago
Isana Yashiro
1,2801622
But the females are not adjacent
– Muhamad Abdul Rosid
49 mins ago
@MuhamadAbdulRosid: That's considered in my answer.
– Isana Yashiro
44 mins ago
I know there are 9 ghosts. But, why is it C(15,6) why 15? I don't understand. Then why C(16,4)? Where does it come from? The 16
– Muhamad Abdul Rosid
11 mins ago
@MuhamadAbdulRosid: Correct, that's what want you to think, that's the key idea/value of this problem, you're on the right path, I will edit for clarification.
– Isana Yashiro
9 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Here is one way of thinking it: First seat the men then seat the women. And arrange people in the right order before seating.
beginalignP(15,6)cdot P(16,4)\=boxed15choose66!cdot16choose44!endalign
It may sounds creepy, but here is another hint if you don't get it:
beginaligntextrmBeyond those people, there are identical 9 ghost there(to help you)endalign
The key idea is about the $15$ and $16$:
- 15: this comes from 19-4, why? because I first consider those 6 men and 9 ghosts.
- 16: now we finish the queuing of 6+9=15 men-ghost-queue, what's the next step to make girls no adjacent?
answered 51 mins ago
Isana Yashiro
1,2801622
Here is one way of thinking it: First seat the men then seat the women. And arrange people in the right order before seating.
beginalignP(15,6)cdot P(16,4)\=boxed15choose66!cdot16choose44!endalign
It may sounds creepy, but here is another hint if you don't get it:
beginaligntextrmBeyond those people, there are identical 9 ghost there(to help you)endalign
The key idea is about the $15$ and $16$:
- 15: this comes from 19-4, why? because I first consider those 6 men and 9 ghosts.
- 16: now we finish the queuing of 6+9=15 men-ghost-queue, what's the next step to make girls no adjacent?
answered 51 mins ago
Isana Yashiro
1,2801622
edited 5 mins ago
answered 51 mins ago
Isana Yashiro
1,2801622
answered 51 mins ago
Isana Yashiro
1,2801622
answered 51 mins ago
Isana Yashiro
1,2801622
1,2801622
But the females are not adjacent
– Muhamad Abdul Rosid
49 mins ago
@MuhamadAbdulRosid: That's considered in my answer.
– Isana Yashiro
44 mins ago
I know there are 9 ghosts. But, why is it C(15,6) why 15? I don't understand. Then why C(16,4)? Where does it come from? The 16
– Muhamad Abdul Rosid
11 mins ago
@MuhamadAbdulRosid: Correct, that's what want you to think, that's the key idea/value of this problem, you're on the right path, I will edit for clarification.
– Isana Yashiro
9 mins ago
add a comment |Â
But the females are not adjacent
– Muhamad Abdul Rosid
49 mins ago
@MuhamadAbdulRosid: That's considered in my answer.
– Isana Yashiro
44 mins ago
I know there are 9 ghosts. But, why is it C(15,6) why 15? I don't understand. Then why C(16,4)? Where does it come from? The 16
– Muhamad Abdul Rosid
11 mins ago
@MuhamadAbdulRosid: Correct, that's what want you to think, that's the key idea/value of this problem, you're on the right path, I will edit for clarification.
– Isana Yashiro
9 mins ago
But the females are not adjacent
– Muhamad Abdul Rosid
49 mins ago
But the females are not adjacent
– Muhamad Abdul Rosid
49 mins ago
@MuhamadAbdulRosid: That's considered in my answer.
– Isana Yashiro
44 mins ago
@MuhamadAbdulRosid: That's considered in my answer.
– Isana Yashiro
44 mins ago
I know there are 9 ghosts. But, why is it C(15,6) why 15? I don't understand. Then why C(16,4)? Where does it come from? The 16
– Muhamad Abdul Rosid
11 mins ago
I know there are 9 ghosts. But, why is it C(15,6) why 15? I don't understand. Then why C(16,4)? Where does it come from? The 16
– Muhamad Abdul Rosid
11 mins ago
@MuhamadAbdulRosid: Correct, that's what want you to think, that's the key idea/value of this problem, you're on the right path, I will edit for clarification.
– Isana Yashiro
9 mins ago
@MuhamadAbdulRosid: Correct, that's what want you to think, that's the key idea/value of this problem, you're on the right path, I will edit for clarification.
– Isana Yashiro
9 mins ago
add a comment |Â
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So there are a lot of empty seats. For the first part, just place the women, so $binom 194$, and then place the men, $binom 156$. (and then assign people). Can you finish from here?
– lulu
2 hours ago
But, the females are not adjacent.
– Muhamad Abdul Rosid
50 mins ago
Do you have the answer? Compare it with 157405248000, if it's so, trust my answer.
– Isana Yashiro
40 mins ago