Selecting a security team that includes at most one of the oldest man or oldest woman
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A security company consists of $7$ men and $6$ women. Determine the number of ways to form the night shift team of $3$ men and $3$ women if it can include at most one of the oldest man or oldest woman.
I know the first step is to pick $3$ from each group which is
$_6C_3 cdot _7C_3 $
but I am not sure how to account for the "include at most one of the oldest man or oldest woman"
The answer is $_6C_3 cdot _7C_3 + _6C_3 cdot _5C_3 + _6C_2 cdot _5C_3 $
Also, how do I account for it in my calculations if I want to include the oldest man and oldest woman from the group?
combinatorics combinations
edited Nov 26 at 11:49
N. F. Taussig
43k93254
asked Nov 26 at 10:37
Erikien
494
|
show 1 more comment
up vote
0
down vote
favorite
A security company consists of $7$ men and $6$ women. Determine the number of ways to form the night shift team of $3$ men and $3$ women if it can include at most one of the oldest man or oldest woman.
I know the first step is to pick $3$ from each group which is
$_6C_3 cdot _7C_3 $
but I am not sure how to account for the "include at most one of the oldest man or oldest woman"
The answer is $_6C_3 cdot _7C_3 + _6C_3 cdot _5C_3 + _6C_2 cdot _5C_3 $
Also, how do I account for it in my calculations if I want to include the oldest man and oldest woman from the group?
combinatorics combinations
edited Nov 26 at 11:49
N. F. Taussig
43k93254
asked Nov 26 at 10:37
Erikien
494
The question in the header doesn't match the one in the body. In the header you say "at least" and in the body you say "at most". Which one did you intend?
– lulu
Nov 26 at 10:49
2
If "at most" , break it into cases. Count those with neither, then those with just the old man, then those with just the old woman.
– lulu
Nov 26 at 10:50
@lulu im not too sure how to account for those with just the old man and those with the old woman. isn't it the same as $6C3 cdot 7C3$
– Erikien
Nov 26 at 11:02
2
For just the old man, you need to choose $2$ men out of the remaining $6$ and $3$ women from the remaining $5$ (five, since you can't choose the eldest woman).
– lulu
Nov 26 at 11:03
The stated answer is not correct. The first term includes all possible selections, including those that contain the oldest man and oldest woman.
– N. F. Taussig
Nov 26 at 11:52
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A security company consists of $7$ men and $6$ women. Determine the number of ways to form the night shift team of $3$ men and $3$ women if it can include at most one of the oldest man or oldest woman.
I know the first step is to pick $3$ from each group which is
$_6C_3 cdot _7C_3 $
but I am not sure how to account for the "include at most one of the oldest man or oldest woman"
The answer is $_6C_3 cdot _7C_3 + _6C_3 cdot _5C_3 + _6C_2 cdot _5C_3 $
Also, how do I account for it in my calculations if I want to include the oldest man and oldest woman from the group?
combinatorics combinations
edited Nov 26 at 11:49
N. F. Taussig
43k93254
asked Nov 26 at 10:37
Erikien
494
A security company consists of $7$ men and $6$ women. Determine the number of ways to form the night shift team of $3$ men and $3$ women if it can include at most one of the oldest man or oldest woman.
I know the first step is to pick $3$ from each group which is
$_6C_3 cdot _7C_3 $
but I am not sure how to account for the "include at most one of the oldest man or oldest woman"
The answer is $_6C_3 cdot _7C_3 + _6C_3 cdot _5C_3 + _6C_2 cdot _5C_3 $
Also, how do I account for it in my calculations if I want to include the oldest man and oldest woman from the group?
combinatorics combinations
combinatorics combinations
edited Nov 26 at 11:49
N. F. Taussig
43k93254
asked Nov 26 at 10:37
Erikien
494
edited Nov 26 at 11:49
N. F. Taussig
43k93254
asked Nov 26 at 10:37
Erikien
494
edited Nov 26 at 11:49
N. F. Taussig
43k93254
edited Nov 26 at 11:49
N. F. Taussig
43k93254
edited Nov 26 at 11:49
N. F. Taussig
43k93254
43k93254
asked Nov 26 at 10:37
Erikien
494
asked Nov 26 at 10:37
Erikien
494
asked Nov 26 at 10:37
Erikien
494
494
The question in the header doesn't match the one in the body. In the header you say "at least" and in the body you say "at most". Which one did you intend?
– lulu
Nov 26 at 10:49
2
If "at most" , break it into cases. Count those with neither, then those with just the old man, then those with just the old woman.
– lulu
Nov 26 at 10:50
@lulu im not too sure how to account for those with just the old man and those with the old woman. isn't it the same as $6C3 cdot 7C3$
– Erikien
Nov 26 at 11:02
2
For just the old man, you need to choose $2$ men out of the remaining $6$ and $3$ women from the remaining $5$ (five, since you can't choose the eldest woman).
– lulu
Nov 26 at 11:03
The stated answer is not correct. The first term includes all possible selections, including those that contain the oldest man and oldest woman.
– N. F. Taussig
Nov 26 at 11:52
|
show 1 more comment
The question in the header doesn't match the one in the body. In the header you say "at least" and in the body you say "at most". Which one did you intend?
– lulu
Nov 26 at 10:49
2
If "at most" , break it into cases. Count those with neither, then those with just the old man, then those with just the old woman.
– lulu
Nov 26 at 10:50
@lulu im not too sure how to account for those with just the old man and those with the old woman. isn't it the same as $6C3 cdot 7C3$
– Erikien
Nov 26 at 11:02
2
For just the old man, you need to choose $2$ men out of the remaining $6$ and $3$ women from the remaining $5$ (five, since you can't choose the eldest woman).
– lulu
Nov 26 at 11:03
The stated answer is not correct. The first term includes all possible selections, including those that contain the oldest man and oldest woman.
– N. F. Taussig
Nov 26 at 11:52
The question in the header doesn't match the one in the body. In the header you say "at least" and in the body you say "at most". Which one did you intend?
– lulu
Nov 26 at 10:49
The question in the header doesn't match the one in the body. In the header you say "at least" and in the body you say "at most". Which one did you intend?
– lulu
Nov 26 at 10:49
2
2
If "at most" , break it into cases. Count those with neither, then those with just the old man, then those with just the old woman.
– lulu
Nov 26 at 10:50
If "at most" , break it into cases. Count those with neither, then those with just the old man, then those with just the old woman.
– lulu
Nov 26 at 10:50
@lulu im not too sure how to account for those with just the old man and those with the old woman. isn't it the same as $6C3 cdot 7C3$
– Erikien
Nov 26 at 11:02
@lulu im not too sure how to account for those with just the old man and those with the old woman. isn't it the same as $6C3 cdot 7C3$
– Erikien
Nov 26 at 11:02
2
2
For just the old man, you need to choose $2$ men out of the remaining $6$ and $3$ women from the remaining $5$ (five, since you can't choose the eldest woman).
– lulu
Nov 26 at 11:03
For just the old man, you need to choose $2$ men out of the remaining $6$ and $3$ women from the remaining $5$ (five, since you can't choose the eldest woman).
– lulu
Nov 26 at 11:03
The stated answer is not correct. The first term includes all possible selections, including those that contain the oldest man and oldest woman.
– N. F. Taussig
Nov 26 at 11:52
The stated answer is not correct. The first term includes all possible selections, including those that contain the oldest man and oldest woman.
– N. F. Taussig
Nov 26 at 11:52
|
show 1 more comment
1 Answer
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oldest
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1
down vote
Method 1: There are three admissible ways to select the teams:
- The oldest man is included, but the oldest woman is not.
- The oldest man is included, but the oldest man is not.
- Neither the oldest man nor the oldest woman is included.
The oldest man is included, but the oldest woman is not: We must select the oldest man, two of the other six men, and three of the six women other than the oldest woman. We can do this in
$$binom{1}{1}binom{6}{2}binom{1}{0}binom{5}{3} = binom{6}{2}binom{5}{3}$$
ways.
The oldest woman is included, but the oldest man is not: We must select the oldest man, two of the other five women, and three of the six men other than the oldest man. We can do this in
$$binom{1}{0}binom{6}{3}binom{1}{1}binom{5}{2} = binom{6}{3}binom{5}{2}$$
ways.
Neither the oldest man nor the oldest woman is included: We must select three of the six men other than the oldest man and three of the five women other than the oldest women. We can do this in
$$binom{1}{0}binom{6}{3}binom{1}{0}binom{5}{3} = binom{6}{3}binom{5}{3}$$
ways.
Total: Since these cases are mutually exclusive and exhaustive, the number of ways a security team of three men and three women can be selected from a team of seven men and six women so that it contains at most one of the oldest man or the oldest woman is
$$binom{6}{2}binom{5}{3} + binom{6}{3}binom{5}{2} + binom{6}{3}binom{5}{3}$$
Method 3: We subtract the number of teams that include most the oldest man and the oldest woman from the total number of teams.
If there were no restrictions, we could choose any three of the seven men and any three of the six women on the team in
$$binom{7}{3}binom{6}{3}$$
ways.
If a team were to include both the oldest man and the oldest women, it would also include two of the other six men and two of the other five women. There are
$$binom{1}{1}binom{6}{2}binom{1}{1}binom{5}{2} = binom{6}{2}binom{5}{2}$$
such teams.
Hence, the number of admissible selections is
$$binom{7}{3}binom{6}{3} - binom{6}{2}binom{5}{2}$$
answered Nov 26 at 14:25
N. F. Taussig
43k93254
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Method 1: There are three admissible ways to select the teams:
- The oldest man is included, but the oldest woman is not.
- The oldest man is included, but the oldest man is not.
- Neither the oldest man nor the oldest woman is included.
The oldest man is included, but the oldest woman is not: We must select the oldest man, two of the other six men, and three of the six women other than the oldest woman. We can do this in
$$binom{1}{1}binom{6}{2}binom{1}{0}binom{5}{3} = binom{6}{2}binom{5}{3}$$
ways.
The oldest woman is included, but the oldest man is not: We must select the oldest man, two of the other five women, and three of the six men other than the oldest man. We can do this in
$$binom{1}{0}binom{6}{3}binom{1}{1}binom{5}{2} = binom{6}{3}binom{5}{2}$$
ways.
Neither the oldest man nor the oldest woman is included: We must select three of the six men other than the oldest man and three of the five women other than the oldest women. We can do this in
$$binom{1}{0}binom{6}{3}binom{1}{0}binom{5}{3} = binom{6}{3}binom{5}{3}$$
ways.
Total: Since these cases are mutually exclusive and exhaustive, the number of ways a security team of three men and three women can be selected from a team of seven men and six women so that it contains at most one of the oldest man or the oldest woman is
$$binom{6}{2}binom{5}{3} + binom{6}{3}binom{5}{2} + binom{6}{3}binom{5}{3}$$
Method 3: We subtract the number of teams that include most the oldest man and the oldest woman from the total number of teams.
If there were no restrictions, we could choose any three of the seven men and any three of the six women on the team in
$$binom{7}{3}binom{6}{3}$$
ways.
If a team were to include both the oldest man and the oldest women, it would also include two of the other six men and two of the other five women. There are
$$binom{1}{1}binom{6}{2}binom{1}{1}binom{5}{2} = binom{6}{2}binom{5}{2}$$
such teams.
Hence, the number of admissible selections is
$$binom{7}{3}binom{6}{3} - binom{6}{2}binom{5}{2}$$
answered Nov 26 at 14:25
N. F. Taussig
43k93254
add a comment |
up vote
1
down vote
Method 1: There are three admissible ways to select the teams:
- The oldest man is included, but the oldest woman is not.
- The oldest man is included, but the oldest man is not.
- Neither the oldest man nor the oldest woman is included.
The oldest man is included, but the oldest woman is not: We must select the oldest man, two of the other six men, and three of the six women other than the oldest woman. We can do this in
$$binom{1}{1}binom{6}{2}binom{1}{0}binom{5}{3} = binom{6}{2}binom{5}{3}$$
ways.
The oldest woman is included, but the oldest man is not: We must select the oldest man, two of the other five women, and three of the six men other than the oldest man. We can do this in
$$binom{1}{0}binom{6}{3}binom{1}{1}binom{5}{2} = binom{6}{3}binom{5}{2}$$
ways.
Neither the oldest man nor the oldest woman is included: We must select three of the six men other than the oldest man and three of the five women other than the oldest women. We can do this in
$$binom{1}{0}binom{6}{3}binom{1}{0}binom{5}{3} = binom{6}{3}binom{5}{3}$$
ways.
Total: Since these cases are mutually exclusive and exhaustive, the number of ways a security team of three men and three women can be selected from a team of seven men and six women so that it contains at most one of the oldest man or the oldest woman is
$$binom{6}{2}binom{5}{3} + binom{6}{3}binom{5}{2} + binom{6}{3}binom{5}{3}$$
Method 3: We subtract the number of teams that include most the oldest man and the oldest woman from the total number of teams.
If there were no restrictions, we could choose any three of the seven men and any three of the six women on the team in
$$binom{7}{3}binom{6}{3}$$
ways.
If a team were to include both the oldest man and the oldest women, it would also include two of the other six men and two of the other five women. There are
$$binom{1}{1}binom{6}{2}binom{1}{1}binom{5}{2} = binom{6}{2}binom{5}{2}$$
such teams.
Hence, the number of admissible selections is
$$binom{7}{3}binom{6}{3} - binom{6}{2}binom{5}{2}$$
answered Nov 26 at 14:25
N. F. Taussig
43k93254
add a comment |
up vote
1
down vote
up vote
1
down vote
Method 1: There are three admissible ways to select the teams:
- The oldest man is included, but the oldest woman is not.
- The oldest man is included, but the oldest man is not.
- Neither the oldest man nor the oldest woman is included.
The oldest man is included, but the oldest woman is not: We must select the oldest man, two of the other six men, and three of the six women other than the oldest woman. We can do this in
$$binom{1}{1}binom{6}{2}binom{1}{0}binom{5}{3} = binom{6}{2}binom{5}{3}$$
ways.
The oldest woman is included, but the oldest man is not: We must select the oldest man, two of the other five women, and three of the six men other than the oldest man. We can do this in
$$binom{1}{0}binom{6}{3}binom{1}{1}binom{5}{2} = binom{6}{3}binom{5}{2}$$
ways.
Neither the oldest man nor the oldest woman is included: We must select three of the six men other than the oldest man and three of the five women other than the oldest women. We can do this in
$$binom{1}{0}binom{6}{3}binom{1}{0}binom{5}{3} = binom{6}{3}binom{5}{3}$$
ways.
Total: Since these cases are mutually exclusive and exhaustive, the number of ways a security team of three men and three women can be selected from a team of seven men and six women so that it contains at most one of the oldest man or the oldest woman is
$$binom{6}{2}binom{5}{3} + binom{6}{3}binom{5}{2} + binom{6}{3}binom{5}{3}$$
Method 3: We subtract the number of teams that include most the oldest man and the oldest woman from the total number of teams.
If there were no restrictions, we could choose any three of the seven men and any three of the six women on the team in
$$binom{7}{3}binom{6}{3}$$
ways.
If a team were to include both the oldest man and the oldest women, it would also include two of the other six men and two of the other five women. There are
$$binom{1}{1}binom{6}{2}binom{1}{1}binom{5}{2} = binom{6}{2}binom{5}{2}$$
such teams.
Hence, the number of admissible selections is
$$binom{7}{3}binom{6}{3} - binom{6}{2}binom{5}{2}$$
answered Nov 26 at 14:25
N. F. Taussig
43k93254
Method 1: There are three admissible ways to select the teams:
- The oldest man is included, but the oldest woman is not.
- The oldest man is included, but the oldest man is not.
- Neither the oldest man nor the oldest woman is included.
The oldest man is included, but the oldest woman is not: We must select the oldest man, two of the other six men, and three of the six women other than the oldest woman. We can do this in
$$binom{1}{1}binom{6}{2}binom{1}{0}binom{5}{3} = binom{6}{2}binom{5}{3}$$
ways.
The oldest woman is included, but the oldest man is not: We must select the oldest man, two of the other five women, and three of the six men other than the oldest man. We can do this in
$$binom{1}{0}binom{6}{3}binom{1}{1}binom{5}{2} = binom{6}{3}binom{5}{2}$$
ways.
Neither the oldest man nor the oldest woman is included: We must select three of the six men other than the oldest man and three of the five women other than the oldest women. We can do this in
$$binom{1}{0}binom{6}{3}binom{1}{0}binom{5}{3} = binom{6}{3}binom{5}{3}$$
ways.
Total: Since these cases are mutually exclusive and exhaustive, the number of ways a security team of three men and three women can be selected from a team of seven men and six women so that it contains at most one of the oldest man or the oldest woman is
$$binom{6}{2}binom{5}{3} + binom{6}{3}binom{5}{2} + binom{6}{3}binom{5}{3}$$
Method 3: We subtract the number of teams that include most the oldest man and the oldest woman from the total number of teams.
If there were no restrictions, we could choose any three of the seven men and any three of the six women on the team in
$$binom{7}{3}binom{6}{3}$$
ways.
If a team were to include both the oldest man and the oldest women, it would also include two of the other six men and two of the other five women. There are
$$binom{1}{1}binom{6}{2}binom{1}{1}binom{5}{2} = binom{6}{2}binom{5}{2}$$
such teams.
Hence, the number of admissible selections is
$$binom{7}{3}binom{6}{3} - binom{6}{2}binom{5}{2}$$
answered Nov 26 at 14:25
N. F. Taussig
43k93254
answered Nov 26 at 14:25
N. F. Taussig
43k93254
answered Nov 26 at 14:25
N. F. Taussig
43k93254
answered Nov 26 at 14:25
N. F. Taussig
43k93254
43k93254
add a comment |
add a comment |
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The question in the header doesn't match the one in the body. In the header you say "at least" and in the body you say "at most". Which one did you intend?
– lulu
Nov 26 at 10:49
2
If "at most" , break it into cases. Count those with neither, then those with just the old man, then those with just the old woman.
– lulu
Nov 26 at 10:50
@lulu im not too sure how to account for those with just the old man and those with the old woman. isn't it the same as $6C3 cdot 7C3$
– Erikien
Nov 26 at 11:02
2
For just the old man, you need to choose $2$ men out of the remaining $6$ and $3$ women from the remaining $5$ (five, since you can't choose the eldest woman).
– lulu
Nov 26 at 11:03
The stated answer is not correct. The first term includes all possible selections, including those that contain the oldest man and oldest woman.
– N. F. Taussig
Nov 26 at 11:52