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 45 mins ago
ZakuZaku
592
$endgroup$
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 45 mins ago
ZakuZaku
592
$endgroup$
$begingroup$
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
37 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
30 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
12 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 45 mins ago
ZakuZaku
592
$endgroup$
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 45 mins ago
ZakuZaku
592
asked 45 mins ago
ZakuZaku
592
asked 45 mins ago
ZakuZaku
592
asked 45 mins ago
ZakuZaku
592
asked 45 mins ago
ZakuZaku
592
592
$begingroup$
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
37 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
30 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
12 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
37 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
30 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
12 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
37 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
37 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
30 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
30 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
12 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
12 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 24 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 23 mins ago
trancelocationtrancelocation
12.6k1826
$endgroup$
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 36 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.
$endgroup$
$begingroup$
Well, that quite goes against the grain of combinatorics, since you've listed nearly all cases. Try to count without listing. Think lazy!
$endgroup$
– M. Vinay
34 mins ago
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
34 mins ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
13 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 24 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 24 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 24 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 24 mins ago
Austin MohrAustin Mohr
20.5k35098
answered 24 mins ago
Austin MohrAustin Mohr
20.5k35098
answered 24 mins ago
Austin MohrAustin Mohr
20.5k35098
answered 24 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 23 mins ago
trancelocationtrancelocation
12.6k1826
$endgroup$
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 23 mins ago
trancelocationtrancelocation
12.6k1826
$endgroup$
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 23 mins ago
trancelocationtrancelocation
12.6k1826
$endgroup$
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 23 mins ago
trancelocationtrancelocation
12.6k1826
answered 23 mins ago
trancelocationtrancelocation
12.6k1826
answered 23 mins ago
trancelocationtrancelocation
12.6k1826
answered 23 mins ago
trancelocationtrancelocation
12.6k1826
12.6k1826
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 36 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.
$endgroup$
$begingroup$
Well, that quite goes against the grain of combinatorics, since you've listed nearly all cases. Try to count without listing. Think lazy!
$endgroup$
– M. Vinay
34 mins ago
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
34 mins ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
13 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 36 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.
$endgroup$
$begingroup$
Well, that quite goes against the grain of combinatorics, since you've listed nearly all cases. Try to count without listing. Think lazy!
$endgroup$
– M. Vinay
34 mins ago
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
34 mins ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
13 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 36 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 36 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 14 mins ago
answered 36 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 36 mins ago
beefstew2011beefstew2011
687
answered 36 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.
$begingroup$
Well, that quite goes against the grain of combinatorics, since you've listed nearly all cases. Try to count without listing. Think lazy!
$endgroup$
– M. Vinay
34 mins ago
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
34 mins ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
13 mins ago
add a comment |
$begingroup$
Well, that quite goes against the grain of combinatorics, since you've listed nearly all cases. Try to count without listing. Think lazy!
$endgroup$
– M. Vinay
34 mins ago
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
34 mins ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
13 mins ago
$begingroup$
Well, that quite goes against the grain of combinatorics, since you've listed nearly all cases. Try to count without listing. Think lazy!
$endgroup$
– M. Vinay
34 mins ago
$begingroup$
Well, that quite goes against the grain of combinatorics, since you've listed nearly all cases. Try to count without listing. Think lazy!
$endgroup$
– M. Vinay
34 mins ago
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
34 mins ago
$begingroup$
True. I'll delete this.
$endgroup$
– beefstew2011
34 mins ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
13 mins ago
$begingroup$
Undeleted with more general answer.
$endgroup$
– beefstew2011
13 mins ago
add a comment |
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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
37 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
30 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
12 mins ago