Mixing
When do we use ' + ' or ' x ' in a combination question?
$begingroup$
When do we use ' + ' and ' x ' in a combination question?
For example, i have this homework question. An elevator allows a maximum of 5 people per lift. In how many ways can a group of 8 people be lifted in 2 trips to the same floor level?
So, well I thought I can do this. My workings: well the first way i said is $8 c 5$, where any 5 from the 8 people can be lifted. The second way, since there are only 3 people left, they all can be lifted at once, so i said $3c3$. Now, this is where i got stuck.
The answer said is $8 c 5$ x $3c3$ = $56$ x $1$ = $56$. But i disagree, i thought i would be $8 c 5$ + $3c3$ = $56$ + $1$ = $57$. Well, my reason for this is that $3c3$ is still a way.
Can someone please give an explanation of why this is the case? Again, my main question is When do we use ' + ' and ' x ' in a combination question? I lived in australia and might not able to give yous my reply straight away.
combinatorics combinations self-learning
asked Jan 26 at 12:22
Fred WeasleyFred Weasley
16710
$endgroup$
add a comment |
$begingroup$
When do we use ' + ' and ' x ' in a combination question?
For example, i have this homework question. An elevator allows a maximum of 5 people per lift. In how many ways can a group of 8 people be lifted in 2 trips to the same floor level?
So, well I thought I can do this. My workings: well the first way i said is $8 c 5$, where any 5 from the 8 people can be lifted. The second way, since there are only 3 people left, they all can be lifted at once, so i said $3c3$. Now, this is where i got stuck.
The answer said is $8 c 5$ x $3c3$ = $56$ x $1$ = $56$. But i disagree, i thought i would be $8 c 5$ + $3c3$ = $56$ + $1$ = $57$. Well, my reason for this is that $3c3$ is still a way.
Can someone please give an explanation of why this is the case? Again, my main question is When do we use ' + ' and ' x ' in a combination question? I lived in australia and might not able to give yous my reply straight away.
combinatorics combinations self-learning
asked Jan 26 at 12:22
Fred WeasleyFred Weasley
16710
$endgroup$
$begingroup$
What about the 4 and 4 cases?
$endgroup$
– badjohn
Jan 26 at 13:52
add a comment |
$begingroup$
When do we use ' + ' and ' x ' in a combination question?
For example, i have this homework question. An elevator allows a maximum of 5 people per lift. In how many ways can a group of 8 people be lifted in 2 trips to the same floor level?
So, well I thought I can do this. My workings: well the first way i said is $8 c 5$, where any 5 from the 8 people can be lifted. The second way, since there are only 3 people left, they all can be lifted at once, so i said $3c3$. Now, this is where i got stuck.
The answer said is $8 c 5$ x $3c3$ = $56$ x $1$ = $56$. But i disagree, i thought i would be $8 c 5$ + $3c3$ = $56$ + $1$ = $57$. Well, my reason for this is that $3c3$ is still a way.
Can someone please give an explanation of why this is the case? Again, my main question is When do we use ' + ' and ' x ' in a combination question? I lived in australia and might not able to give yous my reply straight away.
combinatorics combinations self-learning
asked Jan 26 at 12:22
Fred WeasleyFred Weasley
16710
$endgroup$
When do we use ' + ' and ' x ' in a combination question?
For example, i have this homework question. An elevator allows a maximum of 5 people per lift. In how many ways can a group of 8 people be lifted in 2 trips to the same floor level?
So, well I thought I can do this. My workings: well the first way i said is $8 c 5$, where any 5 from the 8 people can be lifted. The second way, since there are only 3 people left, they all can be lifted at once, so i said $3c3$. Now, this is where i got stuck.
The answer said is $8 c 5$ x $3c3$ = $56$ x $1$ = $56$. But i disagree, i thought i would be $8 c 5$ + $3c3$ = $56$ + $1$ = $57$. Well, my reason for this is that $3c3$ is still a way.
Can someone please give an explanation of why this is the case? Again, my main question is When do we use ' + ' and ' x ' in a combination question? I lived in australia and might not able to give yous my reply straight away.
combinatorics combinations self-learning
combinatorics combinations self-learning
asked Jan 26 at 12:22
Fred WeasleyFred Weasley
16710
asked Jan 26 at 12:22
Fred WeasleyFred Weasley
16710
asked Jan 26 at 12:22
Fred WeasleyFred Weasley
16710
asked Jan 26 at 12:22
Fred WeasleyFred Weasley
16710
asked Jan 26 at 12:22
Fred WeasleyFred Weasley
16710
16710
$begingroup$
What about the 4 and 4 cases?
$endgroup$
– badjohn
Jan 26 at 13:52
add a comment |
$begingroup$
What about the 4 and 4 cases?
$endgroup$
– badjohn
Jan 26 at 13:52
$begingroup$
What about the 4 and 4 cases?
$endgroup$
– badjohn
Jan 26 at 13:52
$begingroup$
What about the 4 and 4 cases?
$endgroup$
– badjohn
Jan 26 at 13:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Short version: $times$ is for "and", $+$is for "or". In this problem, we choose the five people who go up the first time, and then we choose the three people who go up the second time. We always make both choices, and that's the pattern that gives us multiplication.
A $+$ would come in if it was a choice between two categories - say, I can choose to watch a television channel or listen to a radio station. I'm not going to do both at the same time, so my choices come from the number of TV channels plus the number of radio stations.
What about the 4 and 4 cases?
Ah, didn't notice this the first time around. Yes, that's another valid option for the big picture - which means that both the $+$ and $times$ aspects come into things. We can choose (5 people in the first trip and 3 remaining people in the second trip) or (4 people in the first trip and 4 remaining people in the second trip) or (3 people in the first trip and 5 remaining people in the second trip). The total number of ways is $binom{8}{5}cdot binom{3}{3} + binom{8}{4}cdotbinom{4}{4} + binom{8}{3}cdotbinom{5}{5}$.
answered Jan 26 at 12:45
jmerryjmerry
15.5k1632
$endgroup$
$begingroup$
Thanks! When you said 'I'm not going to do both at the same time' are you referring to mutually exclusive?
$endgroup$
– Fred Weasley
Jan 26 at 23:04
$begingroup$
Yes, mutually exclusive. See also the material just added in my edit.
$endgroup$
– jmerry
Jan 26 at 23:06
add a comment |
$begingroup$
For each of the $binom{8}{5}=frac{8cdot7cdot6cdot5cdot4}{1cdot2cdot3cdot4cdot5}=56$ possibilities to choose the first $5$ people out of $8$, there is just one possibility to choose the remaining $3$ people. So You have $56$ "times" $1$ possibilities. You could also visualize this as a tree with $56$ branches. A standard notation for this using binomial coefficients and $cdot$ as the multiplication symbol would be
$$binom{8}{5}cdotbinom{3}{3}=56cdot 1=56.$$
answered Jan 26 at 12:56
Peter MelechPeter Melech
2,702813
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
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votes
$begingroup$
Short version: $times$ is for "and", $+$is for "or". In this problem, we choose the five people who go up the first time, and then we choose the three people who go up the second time. We always make both choices, and that's the pattern that gives us multiplication.
A $+$ would come in if it was a choice between two categories - say, I can choose to watch a television channel or listen to a radio station. I'm not going to do both at the same time, so my choices come from the number of TV channels plus the number of radio stations.
What about the 4 and 4 cases?
Ah, didn't notice this the first time around. Yes, that's another valid option for the big picture - which means that both the $+$ and $times$ aspects come into things. We can choose (5 people in the first trip and 3 remaining people in the second trip) or (4 people in the first trip and 4 remaining people in the second trip) or (3 people in the first trip and 5 remaining people in the second trip). The total number of ways is $binom{8}{5}cdot binom{3}{3} + binom{8}{4}cdotbinom{4}{4} + binom{8}{3}cdotbinom{5}{5}$.
answered Jan 26 at 12:45
jmerryjmerry
15.5k1632
$endgroup$
$begingroup$
Thanks! When you said 'I'm not going to do both at the same time' are you referring to mutually exclusive?
$endgroup$
– Fred Weasley
Jan 26 at 23:04
$begingroup$
Yes, mutually exclusive. See also the material just added in my edit.
$endgroup$
– jmerry
Jan 26 at 23:06
add a comment |
$begingroup$
Short version: $times$ is for "and", $+$is for "or". In this problem, we choose the five people who go up the first time, and then we choose the three people who go up the second time. We always make both choices, and that's the pattern that gives us multiplication.
A $+$ would come in if it was a choice between two categories - say, I can choose to watch a television channel or listen to a radio station. I'm not going to do both at the same time, so my choices come from the number of TV channels plus the number of radio stations.
What about the 4 and 4 cases?
Ah, didn't notice this the first time around. Yes, that's another valid option for the big picture - which means that both the $+$ and $times$ aspects come into things. We can choose (5 people in the first trip and 3 remaining people in the second trip) or (4 people in the first trip and 4 remaining people in the second trip) or (3 people in the first trip and 5 remaining people in the second trip). The total number of ways is $binom{8}{5}cdot binom{3}{3} + binom{8}{4}cdotbinom{4}{4} + binom{8}{3}cdotbinom{5}{5}$.
answered Jan 26 at 12:45
jmerryjmerry
15.5k1632
$endgroup$
$begingroup$
Thanks! When you said 'I'm not going to do both at the same time' are you referring to mutually exclusive?
$endgroup$
– Fred Weasley
Jan 26 at 23:04
$begingroup$
Yes, mutually exclusive. See also the material just added in my edit.
$endgroup$
– jmerry
Jan 26 at 23:06
add a comment |
$begingroup$
Short version: $times$ is for "and", $+$is for "or". In this problem, we choose the five people who go up the first time, and then we choose the three people who go up the second time. We always make both choices, and that's the pattern that gives us multiplication.
A $+$ would come in if it was a choice between two categories - say, I can choose to watch a television channel or listen to a radio station. I'm not going to do both at the same time, so my choices come from the number of TV channels plus the number of radio stations.
What about the 4 and 4 cases?
Ah, didn't notice this the first time around. Yes, that's another valid option for the big picture - which means that both the $+$ and $times$ aspects come into things. We can choose (5 people in the first trip and 3 remaining people in the second trip) or (4 people in the first trip and 4 remaining people in the second trip) or (3 people in the first trip and 5 remaining people in the second trip). The total number of ways is $binom{8}{5}cdot binom{3}{3} + binom{8}{4}cdotbinom{4}{4} + binom{8}{3}cdotbinom{5}{5}$.
answered Jan 26 at 12:45
jmerryjmerry
15.5k1632
$endgroup$
Short version: $times$ is for "and", $+$is for "or". In this problem, we choose the five people who go up the first time, and then we choose the three people who go up the second time. We always make both choices, and that's the pattern that gives us multiplication.
A $+$ would come in if it was a choice between two categories - say, I can choose to watch a television channel or listen to a radio station. I'm not going to do both at the same time, so my choices come from the number of TV channels plus the number of radio stations.
What about the 4 and 4 cases?
Ah, didn't notice this the first time around. Yes, that's another valid option for the big picture - which means that both the $+$ and $times$ aspects come into things. We can choose (5 people in the first trip and 3 remaining people in the second trip) or (4 people in the first trip and 4 remaining people in the second trip) or (3 people in the first trip and 5 remaining people in the second trip). The total number of ways is $binom{8}{5}cdot binom{3}{3} + binom{8}{4}cdotbinom{4}{4} + binom{8}{3}cdotbinom{5}{5}$.
answered Jan 26 at 12:45
jmerryjmerry
15.5k1632
edited Jan 26 at 23:06
answered Jan 26 at 12:45
jmerryjmerry
15.5k1632
answered Jan 26 at 12:45
jmerryjmerry
15.5k1632
answered Jan 26 at 12:45
jmerryjmerry
15.5k1632
15.5k1632
$begingroup$
Thanks! When you said 'I'm not going to do both at the same time' are you referring to mutually exclusive?
$endgroup$
– Fred Weasley
Jan 26 at 23:04
$begingroup$
Yes, mutually exclusive. See also the material just added in my edit.
$endgroup$
– jmerry
Jan 26 at 23:06
add a comment |
$begingroup$
Thanks! When you said 'I'm not going to do both at the same time' are you referring to mutually exclusive?
$endgroup$
– Fred Weasley
Jan 26 at 23:04
$begingroup$
Yes, mutually exclusive. See also the material just added in my edit.
$endgroup$
– jmerry
Jan 26 at 23:06
$begingroup$
Thanks! When you said 'I'm not going to do both at the same time' are you referring to mutually exclusive?
$endgroup$
– Fred Weasley
Jan 26 at 23:04
$begingroup$
Thanks! When you said 'I'm not going to do both at the same time' are you referring to mutually exclusive?
$endgroup$
– Fred Weasley
Jan 26 at 23:04
$begingroup$
Yes, mutually exclusive. See also the material just added in my edit.
$endgroup$
– jmerry
Jan 26 at 23:06
$begingroup$
Yes, mutually exclusive. See also the material just added in my edit.
$endgroup$
– jmerry
Jan 26 at 23:06
add a comment |
$begingroup$
For each of the $binom{8}{5}=frac{8cdot7cdot6cdot5cdot4}{1cdot2cdot3cdot4cdot5}=56$ possibilities to choose the first $5$ people out of $8$, there is just one possibility to choose the remaining $3$ people. So You have $56$ "times" $1$ possibilities. You could also visualize this as a tree with $56$ branches. A standard notation for this using binomial coefficients and $cdot$ as the multiplication symbol would be
$$binom{8}{5}cdotbinom{3}{3}=56cdot 1=56.$$
answered Jan 26 at 12:56
Peter MelechPeter Melech
2,702813
$endgroup$
add a comment |
$begingroup$
For each of the $binom{8}{5}=frac{8cdot7cdot6cdot5cdot4}{1cdot2cdot3cdot4cdot5}=56$ possibilities to choose the first $5$ people out of $8$, there is just one possibility to choose the remaining $3$ people. So You have $56$ "times" $1$ possibilities. You could also visualize this as a tree with $56$ branches. A standard notation for this using binomial coefficients and $cdot$ as the multiplication symbol would be
$$binom{8}{5}cdotbinom{3}{3}=56cdot 1=56.$$
answered Jan 26 at 12:56
Peter MelechPeter Melech
2,702813
$endgroup$
add a comment |
$begingroup$
For each of the $binom{8}{5}=frac{8cdot7cdot6cdot5cdot4}{1cdot2cdot3cdot4cdot5}=56$ possibilities to choose the first $5$ people out of $8$, there is just one possibility to choose the remaining $3$ people. So You have $56$ "times" $1$ possibilities. You could also visualize this as a tree with $56$ branches. A standard notation for this using binomial coefficients and $cdot$ as the multiplication symbol would be
$$binom{8}{5}cdotbinom{3}{3}=56cdot 1=56.$$
answered Jan 26 at 12:56
Peter MelechPeter Melech
2,702813
$endgroup$
For each of the $binom{8}{5}=frac{8cdot7cdot6cdot5cdot4}{1cdot2cdot3cdot4cdot5}=56$ possibilities to choose the first $5$ people out of $8$, there is just one possibility to choose the remaining $3$ people. So You have $56$ "times" $1$ possibilities. You could also visualize this as a tree with $56$ branches. A standard notation for this using binomial coefficients and $cdot$ as the multiplication symbol would be
$$binom{8}{5}cdotbinom{3}{3}=56cdot 1=56.$$
answered Jan 26 at 12:56
Peter MelechPeter Melech
2,702813
answered Jan 26 at 12:56
Peter MelechPeter Melech
2,702813
answered Jan 26 at 12:56
Peter MelechPeter Melech
2,702813
answered Jan 26 at 12:56
Peter MelechPeter Melech
2,702813
2,702813
add a comment |
add a comment |
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$begingroup$
What about the 4 and 4 cases?
$endgroup$
– badjohn
Jan 26 at 13:52