Mixing
Why is this sum simplified to this value?
$begingroup$
We need to find the number of pairs $i,j$ such that $1le i < j le 90.$ and $j - i > 10$. My approach would be to count the total number of pairs which satisfy the first condition, i.e. $90cdot91/2 - 90$. Then subtract the $10$ ineligible pairs from $i = 1$ to $i = 80$, and then subtract the pairs from $i = 81$ to $i= 89$, which gives $-80cdot10-(9+8+7+..+1) = -845$
In total, this is 3160.
But there is a simpler approach. Namely the sum of integers from 1 to 79 which is $79cdot40$? Why does this work?
combinatorics binomial-coefficients intuition
edited Jan 29 at 3:05
J. W. Tanner
4,0611320
asked Jan 28 at 23:59
Dis-integratingDis-integrating
1,043526
$endgroup$
add a comment |
$begingroup$
We need to find the number of pairs $i,j$ such that $1le i < j le 90.$ and $j - i > 10$. My approach would be to count the total number of pairs which satisfy the first condition, i.e. $90cdot91/2 - 90$. Then subtract the $10$ ineligible pairs from $i = 1$ to $i = 80$, and then subtract the pairs from $i = 81$ to $i= 89$, which gives $-80cdot10-(9+8+7+..+1) = -845$
In total, this is 3160.
But there is a simpler approach. Namely the sum of integers from 1 to 79 which is $79cdot40$? Why does this work?
combinatorics binomial-coefficients intuition
edited Jan 29 at 3:05
J. W. Tanner
4,0611320
asked Jan 28 at 23:59
Dis-integratingDis-integrating
1,043526
$endgroup$
1
$begingroup$
I really wonder about the downvote. OP has given their attempt, and wants to know why another attempt is better. There is nothing wrong with this question.
$endgroup$
– K Split X
Jan 29 at 0:02
$begingroup$
@KSplitX some people are grumpy gatekeepers, or simply exist to be contrary.
$endgroup$
– The Count
Jan 29 at 0:07
$begingroup$
i think it was perhaps a little unclear initially. Hopefully it's a bit cleaner now
$endgroup$
– Dis-integrating
Jan 29 at 0:09
$begingroup$
Ha they removed it
$endgroup$
– K Split X
Jan 29 at 0:26
add a comment |
$begingroup$
We need to find the number of pairs $i,j$ such that $1le i < j le 90.$ and $j - i > 10$. My approach would be to count the total number of pairs which satisfy the first condition, i.e. $90cdot91/2 - 90$. Then subtract the $10$ ineligible pairs from $i = 1$ to $i = 80$, and then subtract the pairs from $i = 81$ to $i= 89$, which gives $-80cdot10-(9+8+7+..+1) = -845$
In total, this is 3160.
But there is a simpler approach. Namely the sum of integers from 1 to 79 which is $79cdot40$? Why does this work?
combinatorics binomial-coefficients intuition
edited Jan 29 at 3:05
J. W. Tanner
4,0611320
asked Jan 28 at 23:59
Dis-integratingDis-integrating
1,043526
$endgroup$
We need to find the number of pairs $i,j$ such that $1le i < j le 90.$ and $j - i > 10$. My approach would be to count the total number of pairs which satisfy the first condition, i.e. $90cdot91/2 - 90$. Then subtract the $10$ ineligible pairs from $i = 1$ to $i = 80$, and then subtract the pairs from $i = 81$ to $i= 89$, which gives $-80cdot10-(9+8+7+..+1) = -845$
In total, this is 3160.
But there is a simpler approach. Namely the sum of integers from 1 to 79 which is $79cdot40$? Why does this work?
combinatorics binomial-coefficients intuition
combinatorics binomial-coefficients intuition
edited Jan 29 at 3:05
J. W. Tanner
4,0611320
asked Jan 28 at 23:59
Dis-integratingDis-integrating
1,043526
edited Jan 29 at 3:05
J. W. Tanner
4,0611320
asked Jan 28 at 23:59
Dis-integratingDis-integrating
1,043526
edited Jan 29 at 3:05
J. W. Tanner
4,0611320
edited Jan 29 at 3:05
J. W. Tanner
4,0611320
edited Jan 29 at 3:05
J. W. Tanner
4,0611320
4,0611320
asked Jan 28 at 23:59
Dis-integratingDis-integrating
1,043526
asked Jan 28 at 23:59
Dis-integratingDis-integrating
1,043526
asked Jan 28 at 23:59
Dis-integratingDis-integrating
1,043526
1,043526
1
$begingroup$
I really wonder about the downvote. OP has given their attempt, and wants to know why another attempt is better. There is nothing wrong with this question.
$endgroup$
– K Split X
Jan 29 at 0:02
$begingroup$
@KSplitX some people are grumpy gatekeepers, or simply exist to be contrary.
$endgroup$
– The Count
Jan 29 at 0:07
$begingroup$
i think it was perhaps a little unclear initially. Hopefully it's a bit cleaner now
$endgroup$
– Dis-integrating
Jan 29 at 0:09
$begingroup$
Ha they removed it
$endgroup$
– K Split X
Jan 29 at 0:26
add a comment |
1
$begingroup$
I really wonder about the downvote. OP has given their attempt, and wants to know why another attempt is better. There is nothing wrong with this question.
$endgroup$
– K Split X
Jan 29 at 0:02
$begingroup$
@KSplitX some people are grumpy gatekeepers, or simply exist to be contrary.
$endgroup$
– The Count
Jan 29 at 0:07
$begingroup$
i think it was perhaps a little unclear initially. Hopefully it's a bit cleaner now
$endgroup$
– Dis-integrating
Jan 29 at 0:09
$begingroup$
Ha they removed it
$endgroup$
– K Split X
Jan 29 at 0:26
1
1
$begingroup$
I really wonder about the downvote. OP has given their attempt, and wants to know why another attempt is better. There is nothing wrong with this question.
$endgroup$
– K Split X
Jan 29 at 0:02
$begingroup$
I really wonder about the downvote. OP has given their attempt, and wants to know why another attempt is better. There is nothing wrong with this question.
$endgroup$
– K Split X
Jan 29 at 0:02
$begingroup$
@KSplitX some people are grumpy gatekeepers, or simply exist to be contrary.
$endgroup$
– The Count
Jan 29 at 0:07
$begingroup$
@KSplitX some people are grumpy gatekeepers, or simply exist to be contrary.
$endgroup$
– The Count
Jan 29 at 0:07
$begingroup$
i think it was perhaps a little unclear initially. Hopefully it's a bit cleaner now
$endgroup$
– Dis-integrating
Jan 29 at 0:09
$begingroup$
i think it was perhaps a little unclear initially. Hopefully it's a bit cleaner now
$endgroup$
– Dis-integrating
Jan 29 at 0:09
$begingroup$
Ha they removed it
$endgroup$
– K Split X
Jan 29 at 0:26
$begingroup$
Ha they removed it
$endgroup$
– K Split X
Jan 29 at 0:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider $j'=j-10.$ Then you looking for all pairs $i,j'$ with $1le i < j' le 80.$ This is like picking a 2-element subset out of the 79-element set ${1,2,dots,80}.$ There are $binom{80}{2} = 79cdot 80 / 2 = 3160$ such subsets.
answered Jan 29 at 0:04
Reiner MartinReiner Martin
3,509414
$endgroup$
$begingroup$
thanks, very neat trick
$endgroup$
– Dis-integrating
Jan 29 at 0:06
$begingroup$
just a careful observation. You need to subtract 80 from $binom{80}{2}$ to ensure that $j'$ is not $i$. That is why we end up with 79 times 80 over 2 instead of 80 times 81 over 2
$endgroup$
– Dis-integrating
Jan 29 at 0:40
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider $j'=j-10.$ Then you looking for all pairs $i,j'$ with $1le i < j' le 80.$ This is like picking a 2-element subset out of the 79-element set ${1,2,dots,80}.$ There are $binom{80}{2} = 79cdot 80 / 2 = 3160$ such subsets.
answered Jan 29 at 0:04
Reiner MartinReiner Martin
3,509414
$endgroup$
$begingroup$
thanks, very neat trick
$endgroup$
– Dis-integrating
Jan 29 at 0:06
$begingroup$
just a careful observation. You need to subtract 80 from $binom{80}{2}$ to ensure that $j'$ is not $i$. That is why we end up with 79 times 80 over 2 instead of 80 times 81 over 2
$endgroup$
– Dis-integrating
Jan 29 at 0:40
add a comment |
$begingroup$
Consider $j'=j-10.$ Then you looking for all pairs $i,j'$ with $1le i < j' le 80.$ This is like picking a 2-element subset out of the 79-element set ${1,2,dots,80}.$ There are $binom{80}{2} = 79cdot 80 / 2 = 3160$ such subsets.
answered Jan 29 at 0:04
Reiner MartinReiner Martin
3,509414
$endgroup$
$begingroup$
thanks, very neat trick
$endgroup$
– Dis-integrating
Jan 29 at 0:06
$begingroup$
just a careful observation. You need to subtract 80 from $binom{80}{2}$ to ensure that $j'$ is not $i$. That is why we end up with 79 times 80 over 2 instead of 80 times 81 over 2
$endgroup$
– Dis-integrating
Jan 29 at 0:40
add a comment |
$begingroup$
Consider $j'=j-10.$ Then you looking for all pairs $i,j'$ with $1le i < j' le 80.$ This is like picking a 2-element subset out of the 79-element set ${1,2,dots,80}.$ There are $binom{80}{2} = 79cdot 80 / 2 = 3160$ such subsets.
answered Jan 29 at 0:04
Reiner MartinReiner Martin
3,509414
$endgroup$
Consider $j'=j-10.$ Then you looking for all pairs $i,j'$ with $1le i < j' le 80.$ This is like picking a 2-element subset out of the 79-element set ${1,2,dots,80}.$ There are $binom{80}{2} = 79cdot 80 / 2 = 3160$ such subsets.
answered Jan 29 at 0:04
Reiner MartinReiner Martin
3,509414
answered Jan 29 at 0:04
Reiner MartinReiner Martin
3,509414
answered Jan 29 at 0:04
Reiner MartinReiner Martin
3,509414
answered Jan 29 at 0:04
Reiner MartinReiner Martin
3,509414
3,509414
$begingroup$
thanks, very neat trick
$endgroup$
– Dis-integrating
Jan 29 at 0:06
$begingroup$
just a careful observation. You need to subtract 80 from $binom{80}{2}$ to ensure that $j'$ is not $i$. That is why we end up with 79 times 80 over 2 instead of 80 times 81 over 2
$endgroup$
– Dis-integrating
Jan 29 at 0:40
add a comment |
$begingroup$
thanks, very neat trick
$endgroup$
– Dis-integrating
Jan 29 at 0:06
$begingroup$
just a careful observation. You need to subtract 80 from $binom{80}{2}$ to ensure that $j'$ is not $i$. That is why we end up with 79 times 80 over 2 instead of 80 times 81 over 2
$endgroup$
– Dis-integrating
Jan 29 at 0:40
$begingroup$
thanks, very neat trick
$endgroup$
– Dis-integrating
Jan 29 at 0:06
$begingroup$
thanks, very neat trick
$endgroup$
– Dis-integrating
Jan 29 at 0:06
$begingroup$
just a careful observation. You need to subtract 80 from $binom{80}{2}$ to ensure that $j'$ is not $i$. That is why we end up with 79 times 80 over 2 instead of 80 times 81 over 2
$endgroup$
– Dis-integrating
Jan 29 at 0:40
$begingroup$
just a careful observation. You need to subtract 80 from $binom{80}{2}$ to ensure that $j'$ is not $i$. That is why we end up with 79 times 80 over 2 instead of 80 times 81 over 2
$endgroup$
– Dis-integrating
Jan 29 at 0:40
add a comment |
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$begingroup$
I really wonder about the downvote. OP has given their attempt, and wants to know why another attempt is better. There is nothing wrong with this question.
$endgroup$
– K Split X
Jan 29 at 0:02
$begingroup$
@KSplitX some people are grumpy gatekeepers, or simply exist to be contrary.
$endgroup$
– The Count
Jan 29 at 0:07
$begingroup$
i think it was perhaps a little unclear initially. Hopefully it's a bit cleaner now
$endgroup$
– Dis-integrating
Jan 29 at 0:09
$begingroup$
Ha they removed it
$endgroup$
– K Split X
Jan 29 at 0:26