Mixing
When to use permutations/combination formulas and how to deal with special cases involving them?
$begingroup$
I really need help with the following problem:
A company contains three managers, two accountants, one CFO, and one
CEO. The roles need to filled from a group of 11 men and 10 women. The
workforce must be composed of 4 men and 3 women. If you choose a
particular 7 employees, changing their positions is considered to
change the team. But a manager is a manager and an accountant is a
accountant, so swapping the two managers, for example, does not change
the company.
(a) How many different teams are possible if each person can play any
position? (b) How many different teams are possible if 3 of the 11 men
only know how to be CEOs? (c) How many different teams are possible if
each person can be any role, but Jessica, Marnie (women) and Ryan
(man) cannot all be chosen because they talk too much when placed on
the same team. (Two of the three may be placed on the work team, but
not all three at once).
I really have no idea where to start with this problem. My first instinct for part (a) was to add C(11,4) + C(10,3) = 450 combinations. But that number seems too low. What about the rest? Sorry if this is obvious I'm just brand new to this type of math.
probability combinatorics discrete-mathematics
asked Jan 16 at 2:16
digiHarmoniousdigiHarmonious
303
$endgroup$
add a comment |
$begingroup$
I really need help with the following problem:
A company contains three managers, two accountants, one CFO, and one
CEO. The roles need to filled from a group of 11 men and 10 women. The
workforce must be composed of 4 men and 3 women. If you choose a
particular 7 employees, changing their positions is considered to
change the team. But a manager is a manager and an accountant is a
accountant, so swapping the two managers, for example, does not change
the company.
(a) How many different teams are possible if each person can play any
position? (b) How many different teams are possible if 3 of the 11 men
only know how to be CEOs? (c) How many different teams are possible if
each person can be any role, but Jessica, Marnie (women) and Ryan
(man) cannot all be chosen because they talk too much when placed on
the same team. (Two of the three may be placed on the work team, but
not all three at once).
I really have no idea where to start with this problem. My first instinct for part (a) was to add C(11,4) + C(10,3) = 450 combinations. But that number seems too low. What about the rest? Sorry if this is obvious I'm just brand new to this type of math.
probability combinatorics discrete-mathematics
asked Jan 16 at 2:16
digiHarmoniousdigiHarmonious
303
$endgroup$
1
$begingroup$
Rather than add the ways to choose men and women, you should multiply them (see pwerth's answer). However, after this you're still not quite done, as you need to now assign the $7$ people to the different positions; choose $3$ of them to be a manager, $2$ of them to be accountants, then the remaining spots for CFO and CEO
$endgroup$
– WaveX
Jan 16 at 2:35
$begingroup$
I've updated my answer to account for assigning specific roles, which I originally overlooked in your question.
$endgroup$
– pwerth
Jan 16 at 2:44
add a comment |
$begingroup$
I really need help with the following problem:
A company contains three managers, two accountants, one CFO, and one
CEO. The roles need to filled from a group of 11 men and 10 women. The
workforce must be composed of 4 men and 3 women. If you choose a
particular 7 employees, changing their positions is considered to
change the team. But a manager is a manager and an accountant is a
accountant, so swapping the two managers, for example, does not change
the company.
(a) How many different teams are possible if each person can play any
position? (b) How many different teams are possible if 3 of the 11 men
only know how to be CEOs? (c) How many different teams are possible if
each person can be any role, but Jessica, Marnie (women) and Ryan
(man) cannot all be chosen because they talk too much when placed on
the same team. (Two of the three may be placed on the work team, but
not all three at once).
I really have no idea where to start with this problem. My first instinct for part (a) was to add C(11,4) + C(10,3) = 450 combinations. But that number seems too low. What about the rest? Sorry if this is obvious I'm just brand new to this type of math.
probability combinatorics discrete-mathematics
asked Jan 16 at 2:16
digiHarmoniousdigiHarmonious
303
$endgroup$
I really need help with the following problem:
A company contains three managers, two accountants, one CFO, and one
CEO. The roles need to filled from a group of 11 men and 10 women. The
workforce must be composed of 4 men and 3 women. If you choose a
particular 7 employees, changing their positions is considered to
change the team. But a manager is a manager and an accountant is a
accountant, so swapping the two managers, for example, does not change
the company.
(a) How many different teams are possible if each person can play any
position? (b) How many different teams are possible if 3 of the 11 men
only know how to be CEOs? (c) How many different teams are possible if
each person can be any role, but Jessica, Marnie (women) and Ryan
(man) cannot all be chosen because they talk too much when placed on
the same team. (Two of the three may be placed on the work team, but
not all three at once).
I really have no idea where to start with this problem. My first instinct for part (a) was to add C(11,4) + C(10,3) = 450 combinations. But that number seems too low. What about the rest? Sorry if this is obvious I'm just brand new to this type of math.
probability combinatorics discrete-mathematics
probability combinatorics discrete-mathematics
asked Jan 16 at 2:16
digiHarmoniousdigiHarmonious
303
asked Jan 16 at 2:16
digiHarmoniousdigiHarmonious
303
asked Jan 16 at 2:16
digiHarmoniousdigiHarmonious
303
asked Jan 16 at 2:16
digiHarmoniousdigiHarmonious
303
asked Jan 16 at 2:16
digiHarmoniousdigiHarmonious
303
303
1
$begingroup$
Rather than add the ways to choose men and women, you should multiply them (see pwerth's answer). However, after this you're still not quite done, as you need to now assign the $7$ people to the different positions; choose $3$ of them to be a manager, $2$ of them to be accountants, then the remaining spots for CFO and CEO
$endgroup$
– WaveX
Jan 16 at 2:35
$begingroup$
I've updated my answer to account for assigning specific roles, which I originally overlooked in your question.
$endgroup$
– pwerth
Jan 16 at 2:44
add a comment |
1
$begingroup$
Rather than add the ways to choose men and women, you should multiply them (see pwerth's answer). However, after this you're still not quite done, as you need to now assign the $7$ people to the different positions; choose $3$ of them to be a manager, $2$ of them to be accountants, then the remaining spots for CFO and CEO
$endgroup$
– WaveX
Jan 16 at 2:35
$begingroup$
I've updated my answer to account for assigning specific roles, which I originally overlooked in your question.
$endgroup$
– pwerth
Jan 16 at 2:44
1
1
$begingroup$
Rather than add the ways to choose men and women, you should multiply them (see pwerth's answer). However, after this you're still not quite done, as you need to now assign the $7$ people to the different positions; choose $3$ of them to be a manager, $2$ of them to be accountants, then the remaining spots for CFO and CEO
$endgroup$
– WaveX
Jan 16 at 2:35
$begingroup$
Rather than add the ways to choose men and women, you should multiply them (see pwerth's answer). However, after this you're still not quite done, as you need to now assign the $7$ people to the different positions; choose $3$ of them to be a manager, $2$ of them to be accountants, then the remaining spots for CFO and CEO
$endgroup$
– WaveX
Jan 16 at 2:35
$begingroup$
I've updated my answer to account for assigning specific roles, which I originally overlooked in your question.
$endgroup$
– pwerth
Jan 16 at 2:44
$begingroup$
I've updated my answer to account for assigning specific roles, which I originally overlooked in your question.
$endgroup$
– pwerth
Jan 16 at 2:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For part $(a)$, you want to multiply $binom{11}{4}$ and $binom{10}{3}$, because for every choice of $4$ men, we could have any combination of $3$ women, and vice versa. This is the number of ways to get our $7$ team members. Now, we can assign the managers in $binom{7}{3}$ ways, the accountants in $binom{4}{2}$ ways, and the CFO $binom{2}{1}$ ways. Once we've done this, the CEO is determined. So the answer is
$$binom{11}{4}binom{10}{3}binom{7}{3}binom{4}{2}binom{2}{1}$$
For part $(b)$, observe that either one the CEO is one of those three (in which case he is the only one of those three on the team) or none of them are on the team. If the CEO is one of these three, we have $3$ possible choices for CEO, and $binom{8}{3}$ ways to fill in the other male teammates. So there are $3binom{8}{3}binom{10}{3}$ teams of this form. We can assign the managers in $binom{6}{3}$ ways and the accountants in $binom{3}{2}$ ways.
If the CEO is anyone else, that means these three men are not on the team. In this case, we choose the males in $binom{8}{4}$ ways and the females in $binom{10}{3}$ ways, so there are $binom{8}{4}binom{10}{3}$ teams of this form. We still need to assign roles, so as in part $(a)$ we multiply by $binom{7}{3}binom{4}{2}binom{2}{1}$. All in all, there are
$$3binom{8}{3}binom{10}{3}binom{6}{3}binom{3}{2} +binom{8}{4}binom{10}{3}binom{7}{3}binom{4}{2}binom{2}{1}$$
such teams.
For part $(c)$, we take the total number of possible teams (which is $binom{11}{4}cdotbinom{10}{3}$ from part $(a)$), and subtract the number of teams where Jessica, Marnia and Ryan are all chosen. If these three are given, this means we can choose the remaining male team members in $binom{10}{2}$ ways and the remaining female in $binom{8}{1}$ ways. Accounting for role assignment, the number of such teams is
$$left[binom{11}{4}binom{10}{3} -binom{10}{2}binom{8}{1}right]cdotbinom{7}{3}binom{4}{2}binom{2}{1}$$
answered Jan 16 at 2:32
pwerthpwerth
3,243417
$endgroup$
$begingroup$
Thank you so much! This was extremely helpful :)
$endgroup$
– digiHarmonious
Jan 16 at 3:28
$begingroup$
You're very welcome!
$endgroup$
– pwerth
Jan 16 at 16:13
add a comment |
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$begingroup$
For part $(a)$, you want to multiply $binom{11}{4}$ and $binom{10}{3}$, because for every choice of $4$ men, we could have any combination of $3$ women, and vice versa. This is the number of ways to get our $7$ team members. Now, we can assign the managers in $binom{7}{3}$ ways, the accountants in $binom{4}{2}$ ways, and the CFO $binom{2}{1}$ ways. Once we've done this, the CEO is determined. So the answer is
$$binom{11}{4}binom{10}{3}binom{7}{3}binom{4}{2}binom{2}{1}$$
For part $(b)$, observe that either one the CEO is one of those three (in which case he is the only one of those three on the team) or none of them are on the team. If the CEO is one of these three, we have $3$ possible choices for CEO, and $binom{8}{3}$ ways to fill in the other male teammates. So there are $3binom{8}{3}binom{10}{3}$ teams of this form. We can assign the managers in $binom{6}{3}$ ways and the accountants in $binom{3}{2}$ ways.
If the CEO is anyone else, that means these three men are not on the team. In this case, we choose the males in $binom{8}{4}$ ways and the females in $binom{10}{3}$ ways, so there are $binom{8}{4}binom{10}{3}$ teams of this form. We still need to assign roles, so as in part $(a)$ we multiply by $binom{7}{3}binom{4}{2}binom{2}{1}$. All in all, there are
$$3binom{8}{3}binom{10}{3}binom{6}{3}binom{3}{2} +binom{8}{4}binom{10}{3}binom{7}{3}binom{4}{2}binom{2}{1}$$
such teams.
For part $(c)$, we take the total number of possible teams (which is $binom{11}{4}cdotbinom{10}{3}$ from part $(a)$), and subtract the number of teams where Jessica, Marnia and Ryan are all chosen. If these three are given, this means we can choose the remaining male team members in $binom{10}{2}$ ways and the remaining female in $binom{8}{1}$ ways. Accounting for role assignment, the number of such teams is
$$left[binom{11}{4}binom{10}{3} -binom{10}{2}binom{8}{1}right]cdotbinom{7}{3}binom{4}{2}binom{2}{1}$$
answered Jan 16 at 2:32
pwerthpwerth
3,243417
$endgroup$
$begingroup$
Thank you so much! This was extremely helpful :)
$endgroup$
– digiHarmonious
Jan 16 at 3:28
$begingroup$
You're very welcome!
$endgroup$
– pwerth
Jan 16 at 16:13
add a comment |
$begingroup$
For part $(a)$, you want to multiply $binom{11}{4}$ and $binom{10}{3}$, because for every choice of $4$ men, we could have any combination of $3$ women, and vice versa. This is the number of ways to get our $7$ team members. Now, we can assign the managers in $binom{7}{3}$ ways, the accountants in $binom{4}{2}$ ways, and the CFO $binom{2}{1}$ ways. Once we've done this, the CEO is determined. So the answer is
$$binom{11}{4}binom{10}{3}binom{7}{3}binom{4}{2}binom{2}{1}$$
For part $(b)$, observe that either one the CEO is one of those three (in which case he is the only one of those three on the team) or none of them are on the team. If the CEO is one of these three, we have $3$ possible choices for CEO, and $binom{8}{3}$ ways to fill in the other male teammates. So there are $3binom{8}{3}binom{10}{3}$ teams of this form. We can assign the managers in $binom{6}{3}$ ways and the accountants in $binom{3}{2}$ ways.
If the CEO is anyone else, that means these three men are not on the team. In this case, we choose the males in $binom{8}{4}$ ways and the females in $binom{10}{3}$ ways, so there are $binom{8}{4}binom{10}{3}$ teams of this form. We still need to assign roles, so as in part $(a)$ we multiply by $binom{7}{3}binom{4}{2}binom{2}{1}$. All in all, there are
$$3binom{8}{3}binom{10}{3}binom{6}{3}binom{3}{2} +binom{8}{4}binom{10}{3}binom{7}{3}binom{4}{2}binom{2}{1}$$
such teams.
For part $(c)$, we take the total number of possible teams (which is $binom{11}{4}cdotbinom{10}{3}$ from part $(a)$), and subtract the number of teams where Jessica, Marnia and Ryan are all chosen. If these three are given, this means we can choose the remaining male team members in $binom{10}{2}$ ways and the remaining female in $binom{8}{1}$ ways. Accounting for role assignment, the number of such teams is
$$left[binom{11}{4}binom{10}{3} -binom{10}{2}binom{8}{1}right]cdotbinom{7}{3}binom{4}{2}binom{2}{1}$$
answered Jan 16 at 2:32
pwerthpwerth
3,243417
$endgroup$
$begingroup$
Thank you so much! This was extremely helpful :)
$endgroup$
– digiHarmonious
Jan 16 at 3:28
$begingroup$
You're very welcome!
$endgroup$
– pwerth
Jan 16 at 16:13
add a comment |
$begingroup$
For part $(a)$, you want to multiply $binom{11}{4}$ and $binom{10}{3}$, because for every choice of $4$ men, we could have any combination of $3$ women, and vice versa. This is the number of ways to get our $7$ team members. Now, we can assign the managers in $binom{7}{3}$ ways, the accountants in $binom{4}{2}$ ways, and the CFO $binom{2}{1}$ ways. Once we've done this, the CEO is determined. So the answer is
$$binom{11}{4}binom{10}{3}binom{7}{3}binom{4}{2}binom{2}{1}$$
For part $(b)$, observe that either one the CEO is one of those three (in which case he is the only one of those three on the team) or none of them are on the team. If the CEO is one of these three, we have $3$ possible choices for CEO, and $binom{8}{3}$ ways to fill in the other male teammates. So there are $3binom{8}{3}binom{10}{3}$ teams of this form. We can assign the managers in $binom{6}{3}$ ways and the accountants in $binom{3}{2}$ ways.
If the CEO is anyone else, that means these three men are not on the team. In this case, we choose the males in $binom{8}{4}$ ways and the females in $binom{10}{3}$ ways, so there are $binom{8}{4}binom{10}{3}$ teams of this form. We still need to assign roles, so as in part $(a)$ we multiply by $binom{7}{3}binom{4}{2}binom{2}{1}$. All in all, there are
$$3binom{8}{3}binom{10}{3}binom{6}{3}binom{3}{2} +binom{8}{4}binom{10}{3}binom{7}{3}binom{4}{2}binom{2}{1}$$
such teams.
For part $(c)$, we take the total number of possible teams (which is $binom{11}{4}cdotbinom{10}{3}$ from part $(a)$), and subtract the number of teams where Jessica, Marnia and Ryan are all chosen. If these three are given, this means we can choose the remaining male team members in $binom{10}{2}$ ways and the remaining female in $binom{8}{1}$ ways. Accounting for role assignment, the number of such teams is
$$left[binom{11}{4}binom{10}{3} -binom{10}{2}binom{8}{1}right]cdotbinom{7}{3}binom{4}{2}binom{2}{1}$$
answered Jan 16 at 2:32
pwerthpwerth
3,243417
$endgroup$
For part $(a)$, you want to multiply $binom{11}{4}$ and $binom{10}{3}$, because for every choice of $4$ men, we could have any combination of $3$ women, and vice versa. This is the number of ways to get our $7$ team members. Now, we can assign the managers in $binom{7}{3}$ ways, the accountants in $binom{4}{2}$ ways, and the CFO $binom{2}{1}$ ways. Once we've done this, the CEO is determined. So the answer is
$$binom{11}{4}binom{10}{3}binom{7}{3}binom{4}{2}binom{2}{1}$$
For part $(b)$, observe that either one the CEO is one of those three (in which case he is the only one of those three on the team) or none of them are on the team. If the CEO is one of these three, we have $3$ possible choices for CEO, and $binom{8}{3}$ ways to fill in the other male teammates. So there are $3binom{8}{3}binom{10}{3}$ teams of this form. We can assign the managers in $binom{6}{3}$ ways and the accountants in $binom{3}{2}$ ways.
If the CEO is anyone else, that means these three men are not on the team. In this case, we choose the males in $binom{8}{4}$ ways and the females in $binom{10}{3}$ ways, so there are $binom{8}{4}binom{10}{3}$ teams of this form. We still need to assign roles, so as in part $(a)$ we multiply by $binom{7}{3}binom{4}{2}binom{2}{1}$. All in all, there are
$$3binom{8}{3}binom{10}{3}binom{6}{3}binom{3}{2} +binom{8}{4}binom{10}{3}binom{7}{3}binom{4}{2}binom{2}{1}$$
such teams.
For part $(c)$, we take the total number of possible teams (which is $binom{11}{4}cdotbinom{10}{3}$ from part $(a)$), and subtract the number of teams where Jessica, Marnia and Ryan are all chosen. If these three are given, this means we can choose the remaining male team members in $binom{10}{2}$ ways and the remaining female in $binom{8}{1}$ ways. Accounting for role assignment, the number of such teams is
$$left[binom{11}{4}binom{10}{3} -binom{10}{2}binom{8}{1}right]cdotbinom{7}{3}binom{4}{2}binom{2}{1}$$
answered Jan 16 at 2:32
pwerthpwerth
3,243417
edited Jan 16 at 2:44
answered Jan 16 at 2:32
pwerthpwerth
3,243417
answered Jan 16 at 2:32
pwerthpwerth
3,243417
answered Jan 16 at 2:32
pwerthpwerth
3,243417
3,243417
$begingroup$
Thank you so much! This was extremely helpful :)
$endgroup$
– digiHarmonious
Jan 16 at 3:28
$begingroup$
You're very welcome!
$endgroup$
– pwerth
Jan 16 at 16:13
add a comment |
$begingroup$
Thank you so much! This was extremely helpful :)
$endgroup$
– digiHarmonious
Jan 16 at 3:28
$begingroup$
You're very welcome!
$endgroup$
– pwerth
Jan 16 at 16:13
$begingroup$
Thank you so much! This was extremely helpful :)
$endgroup$
– digiHarmonious
Jan 16 at 3:28
$begingroup$
Thank you so much! This was extremely helpful :)
$endgroup$
– digiHarmonious
Jan 16 at 3:28
$begingroup$
You're very welcome!
$endgroup$
– pwerth
Jan 16 at 16:13
$begingroup$
You're very welcome!
$endgroup$
– pwerth
Jan 16 at 16:13
add a comment |
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$begingroup$
Rather than add the ways to choose men and women, you should multiply them (see pwerth's answer). However, after this you're still not quite done, as you need to now assign the $7$ people to the different positions; choose $3$ of them to be a manager, $2$ of them to be accountants, then the remaining spots for CFO and CEO
$endgroup$
– WaveX
Jan 16 at 2:35
$begingroup$
I've updated my answer to account for assigning specific roles, which I originally overlooked in your question.
$endgroup$
– pwerth
Jan 16 at 2:44