Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes?counting hands shakePuzzle - In how many pairings can 25 married couples dance when exactly 7 men dance with their own wives?Graph Theory number of handshakes of couplesHandshakes in a partyHow many mixed double pairs can be made from 7 married couples provided that no husband and wife plays in a same set?In how many ways can 10 married couples line up for a photograph if every wife stands next to her husband?How many ways are there to order $n$ women and $n$ men in circleFinding the number of combinations.Round table combinatoricsNumber of handshakes - exclusion apporach
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Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes?
counting hands shakePuzzle - In how many pairings can 25 married couples dance when exactly 7 men dance with their own wives?Graph Theory number of handshakes of couplesHandshakes in a partyHow many mixed double pairs can be made from 7 married couples provided that no husband and wife plays in a same set?In how many ways can 10 married couples line up for a photograph if every wife stands next to her husband?How many ways are there to order $n$ women and $n$ men in circleFinding the number of combinations.Round table combinatoricsNumber of handshakes - exclusion apporach
$begingroup$
My book gave the answer as 24. I thought of it like this:
You have four pairs of couples, so you can think of it as M1W2, M2W2, M3W3, M4W4, where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have 4 * 6 handshakes, but in my answer you are double counting. How do I approach this?
combinatorics
asked 1 hour ago
ZakuZaku
642
$endgroup$
add a comment |
$begingroup$
My book gave the answer as 24. I thought of it like this:
You have four pairs of couples, so you can think of it as M1W2, M2W2, M3W3, M4W4, where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have 4 * 6 handshakes, but in my answer you are double counting. How do I approach this?
combinatorics
asked 1 hour ago
ZakuZaku
642
$endgroup$
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (textHandshakes done by the men)$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
53 mins ago
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
45 mins ago
$begingroup$
I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
$endgroup$
– DanielV
28 mins ago
$begingroup$
Only person #1 has to shake hands 6 times, person #2 has already shaken hands with Person #1, so he only has to shake hands with 5 people. So the answer becomes 6+5+4+3+2+1, or 21. So Yes, I believe 21 is correct, to prevent double counting.
$endgroup$
– Issel
9 mins ago
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
5 secs ago
add a comment |
$begingroup$
My book gave the answer as 24. I thought of it like this:
You have four pairs of couples, so you can think of it as M1W2, M2W2, M3W3, M4W4, where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have 4 * 6 handshakes, but in my answer you are double counting. How do I approach this?
combinatorics
asked 1 hour ago
ZakuZaku
642
$endgroup$
My book gave the answer as 24. I thought of it like this:
You have four pairs of couples, so you can think of it as M1W2, M2W2, M3W3, M4W4, where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have 4 * 6 handshakes, but in my answer you are double counting. How do I approach this?
combinatorics
combinatorics
asked 1 hour ago
ZakuZaku
642
asked 1 hour ago
ZakuZaku
642
asked 1 hour ago
ZakuZaku
642
asked 1 hour ago
ZakuZaku
642
asked 1 hour ago
ZakuZaku
642
642
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (textHandshakes done by the men)$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
53 mins ago
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
45 mins ago
$begingroup$
I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
$endgroup$
– DanielV
28 mins ago
$begingroup$
Only person #1 has to shake hands 6 times, person #2 has already shaken hands with Person #1, so he only has to shake hands with 5 people. So the answer becomes 6+5+4+3+2+1, or 21. So Yes, I believe 21 is correct, to prevent double counting.
$endgroup$
– Issel
9 mins ago
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
5 secs ago
add a comment |
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (textHandshakes done by the men)$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
53 mins ago
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
45 mins ago
$begingroup$
I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
$endgroup$
– DanielV
28 mins ago
$begingroup$
Only person #1 has to shake hands 6 times, person #2 has already shaken hands with Person #1, so he only has to shake hands with 5 people. So the answer becomes 6+5+4+3+2+1, or 21. So Yes, I believe 21 is correct, to prevent double counting.
$endgroup$
– Issel
9 mins ago
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
5 secs ago
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (textHandshakes done by the men)$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
53 mins ago
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (textHandshakes done by the men)$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
53 mins ago
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
45 mins ago
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
45 mins ago
$begingroup$
I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
$endgroup$
– DanielV
28 mins ago
$begingroup$
I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
$endgroup$
– DanielV
28 mins ago
$begingroup$
Only person #1 has to shake hands 6 times, person #2 has already shaken hands with Person #1, so he only has to shake hands with 5 people. So the answer becomes 6+5+4+3+2+1, or 21. So Yes, I believe 21 is correct, to prevent double counting.
$endgroup$
– Issel
9 mins ago
$begingroup$
Only person #1 has to shake hands 6 times, person #2 has already shaken hands with Person #1, so he only has to shake hands with 5 people. So the answer becomes 6+5+4+3+2+1, or 21. So Yes, I believe 21 is correct, to prevent double counting.
$endgroup$
– Issel
9 mins ago
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
5 secs ago
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
5 secs ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom82 = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered 40 mins ago
Austin MohrAustin Mohr
20.5k35098
$endgroup$
add a comment |
$begingroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $colorbluebinom82$
- Number of pairs who do not shake hands: $colorblue4$
It follows:
$$mboxnumber of hand shakes without pairs = colorbluebinom82 - colorblue4 = frac8cdot 72 - 4 = 24$$
answered 39 mins ago
trancelocationtrancelocation
12.7k1826
$endgroup$
add a comment |
$begingroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_i=1^k (4k-4i) = sum_i=1^k4k - sum_i=1^k4i = 4k^2 - 4frack(k+1)2 = 4(k^2 - frack^2+k2) = 4(k^2 - (frack^22 + frack2)) = 4(frack^22-frack2) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered 52 mins ago
beefstew2011beefstew2011
687
New contributor
beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
49 mins ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
29 mins ago
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
6 mins ago
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom82 = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered 40 mins ago
Austin MohrAustin Mohr
20.5k35098
$endgroup$
add a comment |
$begingroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom82 = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered 40 mins ago
Austin MohrAustin Mohr
20.5k35098
$endgroup$
add a comment |
$begingroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom82 = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered 40 mins ago
Austin MohrAustin Mohr
20.5k35098
$endgroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom82 = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered 40 mins ago
Austin MohrAustin Mohr
20.5k35098
answered 40 mins ago
Austin MohrAustin Mohr
20.5k35098
answered 40 mins ago
Austin MohrAustin Mohr
20.5k35098
answered 40 mins ago
Austin MohrAustin Mohr
20.5k35098
20.5k35098
add a comment |
add a comment |
$begingroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $colorbluebinom82$
- Number of pairs who do not shake hands: $colorblue4$
It follows:
$$mboxnumber of hand shakes without pairs = colorbluebinom82 - colorblue4 = frac8cdot 72 - 4 = 24$$
answered 39 mins ago
trancelocationtrancelocation
12.7k1826
$endgroup$
add a comment |
$begingroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $colorbluebinom82$
- Number of pairs who do not shake hands: $colorblue4$
It follows:
$$mboxnumber of hand shakes without pairs = colorbluebinom82 - colorblue4 = frac8cdot 72 - 4 = 24$$
answered 39 mins ago
trancelocationtrancelocation
12.7k1826
$endgroup$
add a comment |
$begingroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $colorbluebinom82$
- Number of pairs who do not shake hands: $colorblue4$
It follows:
$$mboxnumber of hand shakes without pairs = colorbluebinom82 - colorblue4 = frac8cdot 72 - 4 = 24$$
answered 39 mins ago
trancelocationtrancelocation
12.7k1826
$endgroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $colorbluebinom82$
- Number of pairs who do not shake hands: $colorblue4$
It follows:
$$mboxnumber of hand shakes without pairs = colorbluebinom82 - colorblue4 = frac8cdot 72 - 4 = 24$$
answered 39 mins ago
trancelocationtrancelocation
12.7k1826
answered 39 mins ago
trancelocationtrancelocation
12.7k1826
answered 39 mins ago
trancelocationtrancelocation
12.7k1826
answered 39 mins ago
trancelocationtrancelocation
12.7k1826
12.7k1826
add a comment |
add a comment |
$begingroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_i=1^k (4k-4i) = sum_i=1^k4k - sum_i=1^k4i = 4k^2 - 4frack(k+1)2 = 4(k^2 - frack^2+k2) = 4(k^2 - (frack^22 + frack2)) = 4(frack^22-frack2) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered 52 mins ago
beefstew2011beefstew2011
687
New contributor
beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
49 mins ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
29 mins ago
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
6 mins ago
add a comment |
$begingroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_i=1^k (4k-4i) = sum_i=1^k4k - sum_i=1^k4i = 4k^2 - 4frack(k+1)2 = 4(k^2 - frack^2+k2) = 4(k^2 - (frack^22 + frack2)) = 4(frack^22-frack2) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered 52 mins ago
beefstew2011beefstew2011
687
New contributor
beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
49 mins ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
29 mins ago
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
6 mins ago
add a comment |
$begingroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_i=1^k (4k-4i) = sum_i=1^k4k - sum_i=1^k4i = 4k^2 - 4frack(k+1)2 = 4(k^2 - frack^2+k2) = 4(k^2 - (frack^22 + frack2)) = 4(frack^22-frack2) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered 52 mins ago
beefstew2011beefstew2011
687
New contributor
beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_i=1^k (4k-4i) = sum_i=1^k4k - sum_i=1^k4i = 4k^2 - 4frack(k+1)2 = 4(k^2 - frack^2+k2) = 4(k^2 - (frack^22 + frack2)) = 4(frack^22-frack2) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered 52 mins ago
beefstew2011beefstew2011
687
New contributor
beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 29 mins ago
answered 52 mins ago
beefstew2011beefstew2011
687
New contributor
beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 52 mins ago
beefstew2011beefstew2011
687
answered 52 mins ago
beefstew2011beefstew2011
687
687
New contributor
beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
49 mins ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
29 mins ago
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
6 mins ago
add a comment |
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
49 mins ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
29 mins ago
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
6 mins ago
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
49 mins ago
$begingroup$
True. I'll delete this.
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– beefstew2011
49 mins ago
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Undeleted with more general answer.
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– beefstew2011
29 mins ago
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Undeleted with more general answer.
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– beefstew2011
29 mins ago
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Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
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– M. Vinay
6 mins ago
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Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
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– M. Vinay
6 mins ago
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$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (textHandshakes done by the men)$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
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– M. Vinay
53 mins ago
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And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
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– M. Vinay
45 mins ago
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I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
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– DanielV
28 mins ago
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Only person #1 has to shake hands 6 times, person #2 has already shaken hands with Person #1, so he only has to shake hands with 5 people. So the answer becomes 6+5+4+3+2+1, or 21. So Yes, I believe 21 is correct, to prevent double counting.
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– Issel
9 mins ago
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@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
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– M. Vinay
5 secs ago