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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
1 hour 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
1 hour 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
1 hour 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
58 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
48 mins 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
1 hour 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
1 hour 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
1 hour 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
58 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
48 mins 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
1 hour 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
1 hour 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
1 hour 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
58 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
48 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
58 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
58 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
48 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
48 mins 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 1 hour 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 1 hour 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 1 hour 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
1 hour ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
1 hour 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
54 mins ago
add a comment |
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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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
Austin MohrAustin Mohr
20.5k35098
answered 1 hour ago
Austin MohrAustin Mohr
20.5k35098
answered 1 hour ago
Austin MohrAustin Mohr
20.5k35098
answered 1 hour 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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
trancelocationtrancelocation
12.7k1826
answered 1 hour ago
trancelocationtrancelocation
12.7k1826
answered 1 hour ago
trancelocationtrancelocation
12.7k1826
answered 1 hour 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 1 hour 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
1 hour ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
1 hour 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
54 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 1 hour 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
1 hour ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
1 hour 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
54 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 1 hour 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 1 hour 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 1 hour ago
answered 1 hour 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 1 hour ago
beefstew2011beefstew2011
687
answered 1 hour 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
1 hour ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
1 hour 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
54 mins ago
add a comment |
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
1 hour ago
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Undeleted with more general answer.
$endgroup$
– beefstew2011
1 hour 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.
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– M. Vinay
54 mins ago
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True. I'll delete this.
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– beefstew2011
1 hour ago
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
1 hour ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
1 hour ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
1 hour 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
54 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
54 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.
$endgroup$
– M. Vinay
1 hour 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
1 hour 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.
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– DanielV
1 hour 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.
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– Issel
58 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 $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
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– M. Vinay
48 mins ago