How many ways can 200 identical balls be distributed into 40 distinct jars?
$begingroup$
How many ways can $200$ identical balls be distributed into $40$ distinct jars so that number Balls in the first $20$ jars is greater than the number of balls in the last $20$ jars?
I came up with two different solutions to the problem but the answers come out differently. The first solution uses symmetry:
There are $binom{239}{40}$ possible ways to place $200$ balls in $40$ jars. We then subtract all possibilities where the number of balls in the first $20$ jars is equal to the number of balls in the last $20$ jars, which is equal to $binom{119}{20}^2$. We then exploit symmetry (the number of possibilities that there are more balls in the first $20$ jars is equal to the number of possibilities that there are more balls in the last twenty jars). The final answer is
$$frac{binom{239}{40} - binom{119}{20}^2}{2}$$
The second solution uses a sum. It comes out to
$$sum_{k=101}^{200} binom{k+19}{20}binom{219-k}{20}$$
combinatorics
edited 10 hours ago
YuiTo Cheng
2,0532637
asked 15 hours ago
David rossDavid ross
311
New contributor
David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 1 more comment
$begingroup$
How many ways can $200$ identical balls be distributed into $40$ distinct jars so that number Balls in the first $20$ jars is greater than the number of balls in the last $20$ jars?
I came up with two different solutions to the problem but the answers come out differently. The first solution uses symmetry:
There are $binom{239}{40}$ possible ways to place $200$ balls in $40$ jars. We then subtract all possibilities where the number of balls in the first $20$ jars is equal to the number of balls in the last $20$ jars, which is equal to $binom{119}{20}^2$. We then exploit symmetry (the number of possibilities that there are more balls in the first $20$ jars is equal to the number of possibilities that there are more balls in the last twenty jars). The final answer is
$$frac{binom{239}{40} - binom{119}{20}^2}{2}$$
The second solution uses a sum. It comes out to
$$sum_{k=101}^{200} binom{k+19}{20}binom{219-k}{20}$$
combinatorics
edited 10 hours ago
YuiTo Cheng
2,0532637
asked 15 hours ago
David rossDavid ross
311
New contributor
David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
15 hours ago
$begingroup$
Hi, welcome to MSE! Could you explain us, please, what do you mean by "($X$ choose $Y$)"?
$endgroup$
– Yanior Weg
15 hours ago
$begingroup$
@YaniorWeg $C^x_y$ that is I think he means combination.
$endgroup$
– Bijayan Ray
15 hours ago
$begingroup$
Is not the number of possible ways to place 200 balls in 40 jars $binom{239}{39}$?
$endgroup$
– user
14 hours ago
$begingroup$
@user Still waking up. Yes, you are correct.
$endgroup$
– N. F. Taussig
14 hours ago
|
show 1 more comment
$begingroup$
How many ways can $200$ identical balls be distributed into $40$ distinct jars so that number Balls in the first $20$ jars is greater than the number of balls in the last $20$ jars?
I came up with two different solutions to the problem but the answers come out differently. The first solution uses symmetry:
There are $binom{239}{40}$ possible ways to place $200$ balls in $40$ jars. We then subtract all possibilities where the number of balls in the first $20$ jars is equal to the number of balls in the last $20$ jars, which is equal to $binom{119}{20}^2$. We then exploit symmetry (the number of possibilities that there are more balls in the first $20$ jars is equal to the number of possibilities that there are more balls in the last twenty jars). The final answer is
$$frac{binom{239}{40} - binom{119}{20}^2}{2}$$
The second solution uses a sum. It comes out to
$$sum_{k=101}^{200} binom{k+19}{20}binom{219-k}{20}$$
combinatorics
edited 10 hours ago
YuiTo Cheng
2,0532637
asked 15 hours ago
David rossDavid ross
311
New contributor
David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
How many ways can $200$ identical balls be distributed into $40$ distinct jars so that number Balls in the first $20$ jars is greater than the number of balls in the last $20$ jars?
I came up with two different solutions to the problem but the answers come out differently. The first solution uses symmetry:
There are $binom{239}{40}$ possible ways to place $200$ balls in $40$ jars. We then subtract all possibilities where the number of balls in the first $20$ jars is equal to the number of balls in the last $20$ jars, which is equal to $binom{119}{20}^2$. We then exploit symmetry (the number of possibilities that there are more balls in the first $20$ jars is equal to the number of possibilities that there are more balls in the last twenty jars). The final answer is
$$frac{binom{239}{40} - binom{119}{20}^2}{2}$$
The second solution uses a sum. It comes out to
$$sum_{k=101}^{200} binom{k+19}{20}binom{219-k}{20}$$
combinatorics
combinatorics
edited 10 hours ago
YuiTo Cheng
2,0532637
asked 15 hours ago
David rossDavid ross
311
New contributor
David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 10 hours ago
YuiTo Cheng
2,0532637
asked 15 hours ago
David rossDavid ross
311
New contributor
David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 10 hours ago
YuiTo Cheng
2,0532637
edited 10 hours ago
YuiTo Cheng
2,0532637
edited 10 hours ago
YuiTo Cheng
2,0532637
2,0532637
asked 15 hours ago
David rossDavid ross
311
New contributor
David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 15 hours ago
David rossDavid ross
311
asked 15 hours ago
David rossDavid ross
311
311
New contributor
David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
15 hours ago
$begingroup$
Hi, welcome to MSE! Could you explain us, please, what do you mean by "($X$ choose $Y$)"?
$endgroup$
– Yanior Weg
15 hours ago
$begingroup$
@YaniorWeg $C^x_y$ that is I think he means combination.
$endgroup$
– Bijayan Ray
15 hours ago
$begingroup$
Is not the number of possible ways to place 200 balls in 40 jars $binom{239}{39}$?
$endgroup$
– user
14 hours ago
$begingroup$
@user Still waking up. Yes, you are correct.
$endgroup$
– N. F. Taussig
14 hours ago
|
show 1 more comment
1
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
15 hours ago
$begingroup$
Hi, welcome to MSE! Could you explain us, please, what do you mean by "($X$ choose $Y$)"?
$endgroup$
– Yanior Weg
15 hours ago
$begingroup$
@YaniorWeg $C^x_y$ that is I think he means combination.
$endgroup$
– Bijayan Ray
15 hours ago
$begingroup$
Is not the number of possible ways to place 200 balls in 40 jars $binom{239}{39}$?
$endgroup$
– user
14 hours ago
$begingroup$
@user Still waking up. Yes, you are correct.
$endgroup$
– N. F. Taussig
14 hours ago
1
1
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
15 hours ago
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
15 hours ago
$begingroup$
Hi, welcome to MSE! Could you explain us, please, what do you mean by "($X$ choose $Y$)"?
$endgroup$
– Yanior Weg
15 hours ago
$begingroup$
Hi, welcome to MSE! Could you explain us, please, what do you mean by "($X$ choose $Y$)"?
$endgroup$
– Yanior Weg
15 hours ago
$begingroup$
@YaniorWeg $C^x_y$ that is I think he means combination.
$endgroup$
– Bijayan Ray
15 hours ago
$begingroup$
@YaniorWeg $C^x_y$ that is I think he means combination.
$endgroup$
– Bijayan Ray
15 hours ago
$begingroup$
Is not the number of possible ways to place 200 balls in 40 jars $binom{239}{39}$?
$endgroup$
– user
14 hours ago
$begingroup$
Is not the number of possible ways to place 200 balls in 40 jars $binom{239}{39}$?
$endgroup$
– user
14 hours ago
$begingroup$
@user Still waking up. Yes, you are correct.
$endgroup$
– N. F. Taussig
14 hours ago
$begingroup$
@user Still waking up. Yes, you are correct.
$endgroup$
– N. F. Taussig
14 hours ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
An expression for the number found by means of stars and bars is:
$$sum_{k=0}^{99}binom{k+19}{19}binom{200-k+19}{19}$$
We can rewrite this as:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}$$under the convention that $binom{n}{m}=0$ if $mnotin{0,1,dots,n}$.
Further we have:$$binom{239}{39}=sum_{i+j=238}binom{i}{19}binom{j}{19}=$$$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}+sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}+binom{119}{19}^2$$where the first equality can be recognized as the hockey-stick equality.
This with:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}$$
so that:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=frac{1}{2}left[binom{239}{39}-binom{119}{19}^2right]$$
answered 14 hours ago
drhabdrhab
103k545136
$endgroup$
$begingroup$
Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
$endgroup$
– Eric Duminil
13 hours ago
add a comment |
$begingroup$
You made an error in applying "stars and bars". You should replace $binom{239}{40}$ with $binom{239}{39}$, and $binom{119}{20}$ with $binom{119}{19}$ and so on.
With correct expressions you obtain:
$$
sum_{k=101}^{200} binom{k+19}{19}binom{219-k}{19}=frac{binom{239}{39} - binom{119}{19}^2}{2}.
$$
No contradiction appears.
answered 14 hours ago
useruser
5,41411030
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
oldest
votes
$begingroup$
An expression for the number found by means of stars and bars is:
$$sum_{k=0}^{99}binom{k+19}{19}binom{200-k+19}{19}$$
We can rewrite this as:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}$$under the convention that $binom{n}{m}=0$ if $mnotin{0,1,dots,n}$.
Further we have:$$binom{239}{39}=sum_{i+j=238}binom{i}{19}binom{j}{19}=$$$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}+sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}+binom{119}{19}^2$$where the first equality can be recognized as the hockey-stick equality.
This with:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}$$
so that:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=frac{1}{2}left[binom{239}{39}-binom{119}{19}^2right]$$
answered 14 hours ago
drhabdrhab
103k545136
$endgroup$
$begingroup$
Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
$endgroup$
– Eric Duminil
13 hours ago
add a comment |
$begingroup$
An expression for the number found by means of stars and bars is:
$$sum_{k=0}^{99}binom{k+19}{19}binom{200-k+19}{19}$$
We can rewrite this as:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}$$under the convention that $binom{n}{m}=0$ if $mnotin{0,1,dots,n}$.
Further we have:$$binom{239}{39}=sum_{i+j=238}binom{i}{19}binom{j}{19}=$$$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}+sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}+binom{119}{19}^2$$where the first equality can be recognized as the hockey-stick equality.
This with:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}$$
so that:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=frac{1}{2}left[binom{239}{39}-binom{119}{19}^2right]$$
answered 14 hours ago
drhabdrhab
103k545136
$endgroup$
$begingroup$
Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
$endgroup$
– Eric Duminil
13 hours ago
add a comment |
$begingroup$
An expression for the number found by means of stars and bars is:
$$sum_{k=0}^{99}binom{k+19}{19}binom{200-k+19}{19}$$
We can rewrite this as:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}$$under the convention that $binom{n}{m}=0$ if $mnotin{0,1,dots,n}$.
Further we have:$$binom{239}{39}=sum_{i+j=238}binom{i}{19}binom{j}{19}=$$$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}+sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}+binom{119}{19}^2$$where the first equality can be recognized as the hockey-stick equality.
This with:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}$$
so that:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=frac{1}{2}left[binom{239}{39}-binom{119}{19}^2right]$$
answered 14 hours ago
drhabdrhab
103k545136
$endgroup$
An expression for the number found by means of stars and bars is:
$$sum_{k=0}^{99}binom{k+19}{19}binom{200-k+19}{19}$$
We can rewrite this as:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}$$under the convention that $binom{n}{m}=0$ if $mnotin{0,1,dots,n}$.
Further we have:$$binom{239}{39}=sum_{i+j=238}binom{i}{19}binom{j}{19}=$$$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}+sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}+binom{119}{19}^2$$where the first equality can be recognized as the hockey-stick equality.
This with:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}$$
so that:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=frac{1}{2}left[binom{239}{39}-binom{119}{19}^2right]$$
answered 14 hours ago
drhabdrhab
103k545136
answered 14 hours ago
drhabdrhab
103k545136
answered 14 hours ago
drhabdrhab
103k545136
answered 14 hours ago
drhabdrhab
103k545136
103k545136
$begingroup$
Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
$endgroup$
– Eric Duminil
13 hours ago
add a comment |
$begingroup$
Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
$endgroup$
– Eric Duminil
13 hours ago
$begingroup$
Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
$endgroup$
– Eric Duminil
13 hours ago
$begingroup$
Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
$endgroup$
– Eric Duminil
13 hours ago
add a comment |
$begingroup$
You made an error in applying "stars and bars". You should replace $binom{239}{40}$ with $binom{239}{39}$, and $binom{119}{20}$ with $binom{119}{19}$ and so on.
With correct expressions you obtain:
$$
sum_{k=101}^{200} binom{k+19}{19}binom{219-k}{19}=frac{binom{239}{39} - binom{119}{19}^2}{2}.
$$
No contradiction appears.
answered 14 hours ago
useruser
5,41411030
$endgroup$
add a comment |
$begingroup$
You made an error in applying "stars and bars". You should replace $binom{239}{40}$ with $binom{239}{39}$, and $binom{119}{20}$ with $binom{119}{19}$ and so on.
With correct expressions you obtain:
$$
sum_{k=101}^{200} binom{k+19}{19}binom{219-k}{19}=frac{binom{239}{39} - binom{119}{19}^2}{2}.
$$
No contradiction appears.
answered 14 hours ago
useruser
5,41411030
$endgroup$
add a comment |
$begingroup$
You made an error in applying "stars and bars". You should replace $binom{239}{40}$ with $binom{239}{39}$, and $binom{119}{20}$ with $binom{119}{19}$ and so on.
With correct expressions you obtain:
$$
sum_{k=101}^{200} binom{k+19}{19}binom{219-k}{19}=frac{binom{239}{39} - binom{119}{19}^2}{2}.
$$
No contradiction appears.
answered 14 hours ago
useruser
5,41411030
$endgroup$
You made an error in applying "stars and bars". You should replace $binom{239}{40}$ with $binom{239}{39}$, and $binom{119}{20}$ with $binom{119}{19}$ and so on.
With correct expressions you obtain:
$$
sum_{k=101}^{200} binom{k+19}{19}binom{219-k}{19}=frac{binom{239}{39} - binom{119}{19}^2}{2}.
$$
No contradiction appears.
answered 14 hours ago
useruser
5,41411030
answered 14 hours ago
useruser
5,41411030
answered 14 hours ago
useruser
5,41411030
answered 14 hours ago
useruser
5,41411030
5,41411030
add a comment |
add a comment |
David ross is a new contributor. Be nice, and check out our Code of Conduct.
David ross is a new contributor. Be nice, and check out our Code of Conduct.
David ross is a new contributor. Be nice, and check out our Code of Conduct.
David ross is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
15 hours ago
$begingroup$
Hi, welcome to MSE! Could you explain us, please, what do you mean by "($X$ choose $Y$)"?
$endgroup$
– Yanior Weg
15 hours ago
$begingroup$
@YaniorWeg $C^x_y$ that is I think he means combination.
$endgroup$
– Bijayan Ray
15 hours ago
$begingroup$
Is not the number of possible ways to place 200 balls in 40 jars $binom{239}{39}$?
$endgroup$
– user
14 hours ago
$begingroup$
@user Still waking up. Yes, you are correct.
$endgroup$
– N. F. Taussig
14 hours ago