Mixing
Rectangular Prisms
Clash Royale CLAN TAG#URR8PPP
up vote
7
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Eight corner bricks are taken out from a 5x5x5 block, which is something like below:
How many rectangular prisms of all sizes can be counted in this block?
Source: Oyun 2018 Final Exam Question
logical-deduction geometry combinatorics
asked Aug 17 at 12:26
Oray
14.1k435139
add a comment |Â
up vote
7
down vote
favorite
Eight corner bricks are taken out from a 5x5x5 block, which is something like below:
How many rectangular prisms of all sizes can be counted in this block?
Source: Oyun 2018 Final Exam Question
logical-deduction geometry combinatorics
asked Aug 17 at 12:26
Oray
14.1k435139
Does a cube count as a rectangular prism??
– Kevin L
Aug 17 at 12:48
1
we dont know if any group of prism creates a cube and since dimensions are not given, yes you will count cubes as prisms.
– Oray
Aug 17 at 12:50
ok gonna start counting :)
– Kevin L
Aug 17 at 12:51
and according to wyzant.com/resources/answers/4393/… cube is also rectangular prism.
– Oray
Aug 17 at 12:52
Since Gareth McCaugham has corrected his slight mistake you should credit him with the accepted answer, for anteriority and for the redaction effort that I just copied and pasted.
– Evargalo
Aug 17 at 20:57
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Eight corner bricks are taken out from a 5x5x5 block, which is something like below:
How many rectangular prisms of all sizes can be counted in this block?
Source: Oyun 2018 Final Exam Question
logical-deduction geometry combinatorics
asked Aug 17 at 12:26
Oray
14.1k435139
Eight corner bricks are taken out from a 5x5x5 block, which is something like below:
How many rectangular prisms of all sizes can be counted in this block?
Source: Oyun 2018 Final Exam Question
logical-deduction geometry combinatorics
asked Aug 17 at 12:26
Oray
14.1k435139
asked Aug 17 at 12:26
Oray
14.1k435139
asked Aug 17 at 12:26
Oray
14.1k435139
asked Aug 17 at 12:26
Oray
14.1k435139
14.1k435139
Does a cube count as a rectangular prism??
– Kevin L
Aug 17 at 12:48
1
we dont know if any group of prism creates a cube and since dimensions are not given, yes you will count cubes as prisms.
– Oray
Aug 17 at 12:50
ok gonna start counting :)
– Kevin L
Aug 17 at 12:51
and according to wyzant.com/resources/answers/4393/… cube is also rectangular prism.
– Oray
Aug 17 at 12:52
Since Gareth McCaugham has corrected his slight mistake you should credit him with the accepted answer, for anteriority and for the redaction effort that I just copied and pasted.
– Evargalo
Aug 17 at 20:57
add a comment |Â
Does a cube count as a rectangular prism??
– Kevin L
Aug 17 at 12:48
1
we dont know if any group of prism creates a cube and since dimensions are not given, yes you will count cubes as prisms.
– Oray
Aug 17 at 12:50
ok gonna start counting :)
– Kevin L
Aug 17 at 12:51
and according to wyzant.com/resources/answers/4393/… cube is also rectangular prism.
– Oray
Aug 17 at 12:52
Since Gareth McCaugham has corrected his slight mistake you should credit him with the accepted answer, for anteriority and for the redaction effort that I just copied and pasted.
– Evargalo
Aug 17 at 20:57
Does a cube count as a rectangular prism??
– Kevin L
Aug 17 at 12:48
Does a cube count as a rectangular prism??
– Kevin L
Aug 17 at 12:48
1
1
we dont know if any group of prism creates a cube and since dimensions are not given, yes you will count cubes as prisms.
– Oray
Aug 17 at 12:50
we dont know if any group of prism creates a cube and since dimensions are not given, yes you will count cubes as prisms.
– Oray
Aug 17 at 12:50
ok gonna start counting :)
– Kevin L
Aug 17 at 12:51
ok gonna start counting :)
– Kevin L
Aug 17 at 12:51
and according to wyzant.com/resources/answers/4393/… cube is also rectangular prism.
– Oray
Aug 17 at 12:52
and according to wyzant.com/resources/answers/4393/… cube is also rectangular prism.
– Oray
Aug 17 at 12:52
Since Gareth McCaugham has corrected his slight mistake you should credit him with the accepted answer, for anteriority and for the redaction effort that I just copied and pasted.
– Evargalo
Aug 17 at 20:57
Since Gareth McCaugham has corrected his slight mistake you should credit him with the accepted answer, for anteriority and for the redaction effort that I just copied and pasted.
– Evargalo
Aug 17 at 20:57
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
Suppose we leave the corners there. Then
a rectangular prism (= cuboid) is defined by three pairs of planes, so there are $binom62^3=15^3=3375$ of these.
How many of these are excluded by having no corners?
A cuboid $[a,b]times[c,d]times[e,f]$ uses a corner iff ($a=0$ OR $b=5$) AND ($c=0$ OR $d=5$) AND ($e=0$ OR $f=5$). The number of ways to choose $(a,b)$ so that this happens is 11, so there are $11^3=1331$ of these.
[EDITED to add:]
Oops, turns out I meant 9 not 11. 01 02 03 04 05 15 25 35 45. So 729 rather than 1331, leading to a correspondingly larger final answer.
So the number of "good" cuboids is
3375 minus 1331 = 2044. [EDITED to add:] Nope, 3375-729 = 2646.
answered Aug 17 at 13:06
Gareth McCaughan♦
54.8k3136213
you're counting a=b=0 etc.
– JonMark Perry
Aug 17 at 13:48
Yes, I am. There are six relevant planes for each axis, from x=0 to x=5 etc. Any way of picking two different ones on each axis yields a cuboid.
– Gareth McCaughan♦
Aug 17 at 15:40
Oh, sorry, you mean I'm counting (0,0) as well as (0,5) etc.? No, I wasn't. I did miscount, but it was a simple off-by-one error rather than anything that interesting.
– Gareth McCaughan♦
Aug 17 at 15:49
it looks like you counted [0,0]x[1,3]x[2,4] for example (in the second part)
– JonMark Perry
Aug 17 at 15:52
I don't think so. I think I just thought "01 02 ... 05 / 05 15 ... 45, one overlap, that'll be 11" when in fact it should have been 9. I'm not sure my thought process was even as conscious as that.
– Gareth McCaughan♦
Aug 17 at 16:45
add a comment |Â
up vote
6
down vote
I think Gareth McCaughan has the right reasoning but a small calculation error. I'll copy his explanations here :
Suppose we leave the corners there. Then
Let's give coordinates to the cubes from 1 to 5 in length, width and height. A rectangular prism (= cuboïd) is defined by its smallest and biggest index (possibly equal) on each axis, so there are $(binom52+5)^3=15^3=3375$ of these.
How many of these are excluded by having no corners?
A cuboïd $[a,b]times[c,d]times[e,f]$ uses a corner iff ($a=1$ OR $b=5$) AND ($c=1$ OR $d=5$) AND ($e=1$ OR $f=5$). The number of ways to choose $(a,b)$ so that this happens is $9$, so there are $9^3=729$ of these.
Alternatively, by inclusion/exclusion:
There are $5 times 5 times 5=125$ cuboïds using the 'vanished' $(1,1,1)$ corner cube, and $8$ ways to pick a corner.
There are $5 times 5=25$ cuboïds using the 'vanished' $(1,1,1)$ and $(1,1,5)$ cubes, and $12$ ways to pick an edge.
There are $5$ cuboïds using the four 'vanished' $(1,1,1)$, $(1,5,1)$, $(1,5,5)$ and $(1,1,5)$ cubes on any given face, and $6$ ways to pick a face.
There is $1$ cuboïd using all eight 'vanished' corner cubes.
The number of 'wrong' cuboïds is then: $(125 times 8)-(25 times 12)+ (5 times 6) -1=729$
So the number of "good" cuboïds is
$3375 - 729 = mathbf2646$.
answered Aug 17 at 13:32
Evargalo
869211
Evargalo, I think you had it initially correct in your alternative explanation; your math no longer tallies in the final equation of your third paragraph.
– El-Guest
Aug 17 at 13:43
@El-Guest. Thanks, I've edited and I hope it is correct now.
– Evargalo
Aug 17 at 13:44
Looks much better. I can clean up some of the math formatting for you as well.
– El-Guest
Aug 17 at 13:46
1
Looks great! thanks again.
– Evargalo
Aug 17 at 13:49
2
This is slightly incorrect because it has "1 or 5" everywhere, where it should be "0 or 5". I confess myself slightly miffed that someone else is getting the coveted green checkmark for copying-and-pasting my answer, fixing a trivial error, and making another trivial error in turn, but never mind. I suppose there's the inclusion-exclusion bit. (I decided not to do it that way because the argument I actually gave was much easier to follow.)
– Gareth McCaughan♦
Aug 17 at 15:47
 |Â
show 2 more comments
up vote
3
down vote
A computer version in JavaScript:
<!doctype html>
<html>
<title>Cuboids</title>
<body>
<span id='out'></span><br>
<button onclick='go();'>go</button>
</body>
<script>
function go() c==14)) d++;
out.textContent=d;
d=0;
for (let a=0;a<5;a++)
for (let b=0;b<5;b++)
for (let c=0;c<5;c++)
d+=countCuboids(a,b,c);
out.textContent+=' :: '+d;
function countCuboids(a,b,c)
if (a%4==0 && b%4==0 && c%4==0) return 0;
if (a%4>0 && b%4==0 && c%4==0) return (4-a)*(5-b)*(5-c);
if (a%4==0 && b%4>0 && c%4==0) return (5-a)*(4-b)*(5-c);
if (a%4==0 && b%4==0 && c%4>0) return (5-a)*(5-b)*(4-c);
if (a%4>0 && b%4>0 && c%4==0) return (5-a)*(5-b)*(5-c)-(5-c);
if (a%4>0 && b%4==0 && c%4>0) return (5-a)*(5-b)*(5-c)-(5-b);
if (a%4==0 && b%4>0 && c%4>0) return (5-a)*(5-b)*(5-c)-(5-a);
if (a%4>0 && b%4>0 && c%4>0) return (5-a)*(5-b)*(5-c)-1;
</script>
</html>
The first version maps $0..14$ to $1,12,123,1234,12345,2,23,234,2345,3,34,345,4,45,5$ and counts anything that has a $1$ or a $5$ in all three loop variables.
Result: 729 (=$15^3-2646$)
The second uses an explicit function to calculate the number of right, up, back cuboids from a starting cube, and sums them.
Result: 2646
answered Aug 17 at 15:14
JonMark Perry
13.3k42666
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Suppose we leave the corners there. Then
a rectangular prism (= cuboid) is defined by three pairs of planes, so there are $binom62^3=15^3=3375$ of these.
How many of these are excluded by having no corners?
A cuboid $[a,b]times[c,d]times[e,f]$ uses a corner iff ($a=0$ OR $b=5$) AND ($c=0$ OR $d=5$) AND ($e=0$ OR $f=5$). The number of ways to choose $(a,b)$ so that this happens is 11, so there are $11^3=1331$ of these.
[EDITED to add:]
Oops, turns out I meant 9 not 11. 01 02 03 04 05 15 25 35 45. So 729 rather than 1331, leading to a correspondingly larger final answer.
So the number of "good" cuboids is
3375 minus 1331 = 2044. [EDITED to add:] Nope, 3375-729 = 2646.
answered Aug 17 at 13:06
Gareth McCaughan♦
54.8k3136213
you're counting a=b=0 etc.
– JonMark Perry
Aug 17 at 13:48
Yes, I am. There are six relevant planes for each axis, from x=0 to x=5 etc. Any way of picking two different ones on each axis yields a cuboid.
– Gareth McCaughan♦
Aug 17 at 15:40
Oh, sorry, you mean I'm counting (0,0) as well as (0,5) etc.? No, I wasn't. I did miscount, but it was a simple off-by-one error rather than anything that interesting.
– Gareth McCaughan♦
Aug 17 at 15:49
it looks like you counted [0,0]x[1,3]x[2,4] for example (in the second part)
– JonMark Perry
Aug 17 at 15:52
I don't think so. I think I just thought "01 02 ... 05 / 05 15 ... 45, one overlap, that'll be 11" when in fact it should have been 9. I'm not sure my thought process was even as conscious as that.
– Gareth McCaughan♦
Aug 17 at 16:45
add a comment |Â
up vote
6
down vote
accepted
Suppose we leave the corners there. Then
a rectangular prism (= cuboid) is defined by three pairs of planes, so there are $binom62^3=15^3=3375$ of these.
How many of these are excluded by having no corners?
A cuboid $[a,b]times[c,d]times[e,f]$ uses a corner iff ($a=0$ OR $b=5$) AND ($c=0$ OR $d=5$) AND ($e=0$ OR $f=5$). The number of ways to choose $(a,b)$ so that this happens is 11, so there are $11^3=1331$ of these.
[EDITED to add:]
Oops, turns out I meant 9 not 11. 01 02 03 04 05 15 25 35 45. So 729 rather than 1331, leading to a correspondingly larger final answer.
So the number of "good" cuboids is
3375 minus 1331 = 2044. [EDITED to add:] Nope, 3375-729 = 2646.
answered Aug 17 at 13:06
Gareth McCaughan♦
54.8k3136213
you're counting a=b=0 etc.
– JonMark Perry
Aug 17 at 13:48
Yes, I am. There are six relevant planes for each axis, from x=0 to x=5 etc. Any way of picking two different ones on each axis yields a cuboid.
– Gareth McCaughan♦
Aug 17 at 15:40
Oh, sorry, you mean I'm counting (0,0) as well as (0,5) etc.? No, I wasn't. I did miscount, but it was a simple off-by-one error rather than anything that interesting.
– Gareth McCaughan♦
Aug 17 at 15:49
it looks like you counted [0,0]x[1,3]x[2,4] for example (in the second part)
– JonMark Perry
Aug 17 at 15:52
I don't think so. I think I just thought "01 02 ... 05 / 05 15 ... 45, one overlap, that'll be 11" when in fact it should have been 9. I'm not sure my thought process was even as conscious as that.
– Gareth McCaughan♦
Aug 17 at 16:45
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Suppose we leave the corners there. Then
a rectangular prism (= cuboid) is defined by three pairs of planes, so there are $binom62^3=15^3=3375$ of these.
How many of these are excluded by having no corners?
A cuboid $[a,b]times[c,d]times[e,f]$ uses a corner iff ($a=0$ OR $b=5$) AND ($c=0$ OR $d=5$) AND ($e=0$ OR $f=5$). The number of ways to choose $(a,b)$ so that this happens is 11, so there are $11^3=1331$ of these.
[EDITED to add:]
Oops, turns out I meant 9 not 11. 01 02 03 04 05 15 25 35 45. So 729 rather than 1331, leading to a correspondingly larger final answer.
So the number of "good" cuboids is
3375 minus 1331 = 2044. [EDITED to add:] Nope, 3375-729 = 2646.
answered Aug 17 at 13:06
Gareth McCaughan♦
54.8k3136213
Suppose we leave the corners there. Then
a rectangular prism (= cuboid) is defined by three pairs of planes, so there are $binom62^3=15^3=3375$ of these.
How many of these are excluded by having no corners?
A cuboid $[a,b]times[c,d]times[e,f]$ uses a corner iff ($a=0$ OR $b=5$) AND ($c=0$ OR $d=5$) AND ($e=0$ OR $f=5$). The number of ways to choose $(a,b)$ so that this happens is 11, so there are $11^3=1331$ of these.
[EDITED to add:]
Oops, turns out I meant 9 not 11. 01 02 03 04 05 15 25 35 45. So 729 rather than 1331, leading to a correspondingly larger final answer.
So the number of "good" cuboids is
3375 minus 1331 = 2044. [EDITED to add:] Nope, 3375-729 = 2646.
answered Aug 17 at 13:06
Gareth McCaughan♦
54.8k3136213
edited Aug 17 at 15:49
answered Aug 17 at 13:06
Gareth McCaughan♦
54.8k3136213
answered Aug 17 at 13:06
Gareth McCaughan♦
54.8k3136213
answered Aug 17 at 13:06
Gareth McCaughan♦
54.8k3136213
54.8k3136213
you're counting a=b=0 etc.
– JonMark Perry
Aug 17 at 13:48
Yes, I am. There are six relevant planes for each axis, from x=0 to x=5 etc. Any way of picking two different ones on each axis yields a cuboid.
– Gareth McCaughan♦
Aug 17 at 15:40
Oh, sorry, you mean I'm counting (0,0) as well as (0,5) etc.? No, I wasn't. I did miscount, but it was a simple off-by-one error rather than anything that interesting.
– Gareth McCaughan♦
Aug 17 at 15:49
it looks like you counted [0,0]x[1,3]x[2,4] for example (in the second part)
– JonMark Perry
Aug 17 at 15:52
I don't think so. I think I just thought "01 02 ... 05 / 05 15 ... 45, one overlap, that'll be 11" when in fact it should have been 9. I'm not sure my thought process was even as conscious as that.
– Gareth McCaughan♦
Aug 17 at 16:45
add a comment |Â
you're counting a=b=0 etc.
– JonMark Perry
Aug 17 at 13:48
Yes, I am. There are six relevant planes for each axis, from x=0 to x=5 etc. Any way of picking two different ones on each axis yields a cuboid.
– Gareth McCaughan♦
Aug 17 at 15:40
Oh, sorry, you mean I'm counting (0,0) as well as (0,5) etc.? No, I wasn't. I did miscount, but it was a simple off-by-one error rather than anything that interesting.
– Gareth McCaughan♦
Aug 17 at 15:49
it looks like you counted [0,0]x[1,3]x[2,4] for example (in the second part)
– JonMark Perry
Aug 17 at 15:52
I don't think so. I think I just thought "01 02 ... 05 / 05 15 ... 45, one overlap, that'll be 11" when in fact it should have been 9. I'm not sure my thought process was even as conscious as that.
– Gareth McCaughan♦
Aug 17 at 16:45
you're counting a=b=0 etc.
– JonMark Perry
Aug 17 at 13:48
you're counting a=b=0 etc.
– JonMark Perry
Aug 17 at 13:48
Yes, I am. There are six relevant planes for each axis, from x=0 to x=5 etc. Any way of picking two different ones on each axis yields a cuboid.
– Gareth McCaughan♦
Aug 17 at 15:40
Yes, I am. There are six relevant planes for each axis, from x=0 to x=5 etc. Any way of picking two different ones on each axis yields a cuboid.
– Gareth McCaughan♦
Aug 17 at 15:40
Oh, sorry, you mean I'm counting (0,0) as well as (0,5) etc.? No, I wasn't. I did miscount, but it was a simple off-by-one error rather than anything that interesting.
– Gareth McCaughan♦
Aug 17 at 15:49
Oh, sorry, you mean I'm counting (0,0) as well as (0,5) etc.? No, I wasn't. I did miscount, but it was a simple off-by-one error rather than anything that interesting.
– Gareth McCaughan♦
Aug 17 at 15:49
it looks like you counted [0,0]x[1,3]x[2,4] for example (in the second part)
– JonMark Perry
Aug 17 at 15:52
it looks like you counted [0,0]x[1,3]x[2,4] for example (in the second part)
– JonMark Perry
Aug 17 at 15:52
I don't think so. I think I just thought "01 02 ... 05 / 05 15 ... 45, one overlap, that'll be 11" when in fact it should have been 9. I'm not sure my thought process was even as conscious as that.
– Gareth McCaughan♦
Aug 17 at 16:45
I don't think so. I think I just thought "01 02 ... 05 / 05 15 ... 45, one overlap, that'll be 11" when in fact it should have been 9. I'm not sure my thought process was even as conscious as that.
– Gareth McCaughan♦
Aug 17 at 16:45
add a comment |Â
up vote
6
down vote
I think Gareth McCaughan has the right reasoning but a small calculation error. I'll copy his explanations here :
Suppose we leave the corners there. Then
Let's give coordinates to the cubes from 1 to 5 in length, width and height. A rectangular prism (= cuboïd) is defined by its smallest and biggest index (possibly equal) on each axis, so there are $(binom52+5)^3=15^3=3375$ of these.
How many of these are excluded by having no corners?
A cuboïd $[a,b]times[c,d]times[e,f]$ uses a corner iff ($a=1$ OR $b=5$) AND ($c=1$ OR $d=5$) AND ($e=1$ OR $f=5$). The number of ways to choose $(a,b)$ so that this happens is $9$, so there are $9^3=729$ of these.
Alternatively, by inclusion/exclusion:
There are $5 times 5 times 5=125$ cuboïds using the 'vanished' $(1,1,1)$ corner cube, and $8$ ways to pick a corner.
There are $5 times 5=25$ cuboïds using the 'vanished' $(1,1,1)$ and $(1,1,5)$ cubes, and $12$ ways to pick an edge.
There are $5$ cuboïds using the four 'vanished' $(1,1,1)$, $(1,5,1)$, $(1,5,5)$ and $(1,1,5)$ cubes on any given face, and $6$ ways to pick a face.
There is $1$ cuboïd using all eight 'vanished' corner cubes.
The number of 'wrong' cuboïds is then: $(125 times 8)-(25 times 12)+ (5 times 6) -1=729$
So the number of "good" cuboïds is
$3375 - 729 = mathbf2646$.
answered Aug 17 at 13:32
Evargalo
869211
Evargalo, I think you had it initially correct in your alternative explanation; your math no longer tallies in the final equation of your third paragraph.
– El-Guest
Aug 17 at 13:43
@El-Guest. Thanks, I've edited and I hope it is correct now.
– Evargalo
Aug 17 at 13:44
Looks much better. I can clean up some of the math formatting for you as well.
– El-Guest
Aug 17 at 13:46
1
Looks great! thanks again.
– Evargalo
Aug 17 at 13:49
2
This is slightly incorrect because it has "1 or 5" everywhere, where it should be "0 or 5". I confess myself slightly miffed that someone else is getting the coveted green checkmark for copying-and-pasting my answer, fixing a trivial error, and making another trivial error in turn, but never mind. I suppose there's the inclusion-exclusion bit. (I decided not to do it that way because the argument I actually gave was much easier to follow.)
– Gareth McCaughan♦
Aug 17 at 15:47
 |Â
show 2 more comments
up vote
6
down vote
I think Gareth McCaughan has the right reasoning but a small calculation error. I'll copy his explanations here :
Suppose we leave the corners there. Then
Let's give coordinates to the cubes from 1 to 5 in length, width and height. A rectangular prism (= cuboïd) is defined by its smallest and biggest index (possibly equal) on each axis, so there are $(binom52+5)^3=15^3=3375$ of these.
How many of these are excluded by having no corners?
A cuboïd $[a,b]times[c,d]times[e,f]$ uses a corner iff ($a=1$ OR $b=5$) AND ($c=1$ OR $d=5$) AND ($e=1$ OR $f=5$). The number of ways to choose $(a,b)$ so that this happens is $9$, so there are $9^3=729$ of these.
Alternatively, by inclusion/exclusion:
There are $5 times 5 times 5=125$ cuboïds using the 'vanished' $(1,1,1)$ corner cube, and $8$ ways to pick a corner.
There are $5 times 5=25$ cuboïds using the 'vanished' $(1,1,1)$ and $(1,1,5)$ cubes, and $12$ ways to pick an edge.
There are $5$ cuboïds using the four 'vanished' $(1,1,1)$, $(1,5,1)$, $(1,5,5)$ and $(1,1,5)$ cubes on any given face, and $6$ ways to pick a face.
There is $1$ cuboïd using all eight 'vanished' corner cubes.
The number of 'wrong' cuboïds is then: $(125 times 8)-(25 times 12)+ (5 times 6) -1=729$
So the number of "good" cuboïds is
$3375 - 729 = mathbf2646$.
answered Aug 17 at 13:32
Evargalo
869211
Evargalo, I think you had it initially correct in your alternative explanation; your math no longer tallies in the final equation of your third paragraph.
– El-Guest
Aug 17 at 13:43
@El-Guest. Thanks, I've edited and I hope it is correct now.
– Evargalo
Aug 17 at 13:44
Looks much better. I can clean up some of the math formatting for you as well.
– El-Guest
Aug 17 at 13:46
1
Looks great! thanks again.
– Evargalo
Aug 17 at 13:49
2
This is slightly incorrect because it has "1 or 5" everywhere, where it should be "0 or 5". I confess myself slightly miffed that someone else is getting the coveted green checkmark for copying-and-pasting my answer, fixing a trivial error, and making another trivial error in turn, but never mind. I suppose there's the inclusion-exclusion bit. (I decided not to do it that way because the argument I actually gave was much easier to follow.)
– Gareth McCaughan♦
Aug 17 at 15:47
 |Â
show 2 more comments
up vote
6
down vote
up vote
6
down vote
I think Gareth McCaughan has the right reasoning but a small calculation error. I'll copy his explanations here :
Suppose we leave the corners there. Then
Let's give coordinates to the cubes from 1 to 5 in length, width and height. A rectangular prism (= cuboïd) is defined by its smallest and biggest index (possibly equal) on each axis, so there are $(binom52+5)^3=15^3=3375$ of these.
How many of these are excluded by having no corners?
A cuboïd $[a,b]times[c,d]times[e,f]$ uses a corner iff ($a=1$ OR $b=5$) AND ($c=1$ OR $d=5$) AND ($e=1$ OR $f=5$). The number of ways to choose $(a,b)$ so that this happens is $9$, so there are $9^3=729$ of these.
Alternatively, by inclusion/exclusion:
There are $5 times 5 times 5=125$ cuboïds using the 'vanished' $(1,1,1)$ corner cube, and $8$ ways to pick a corner.
There are $5 times 5=25$ cuboïds using the 'vanished' $(1,1,1)$ and $(1,1,5)$ cubes, and $12$ ways to pick an edge.
There are $5$ cuboïds using the four 'vanished' $(1,1,1)$, $(1,5,1)$, $(1,5,5)$ and $(1,1,5)$ cubes on any given face, and $6$ ways to pick a face.
There is $1$ cuboïd using all eight 'vanished' corner cubes.
The number of 'wrong' cuboïds is then: $(125 times 8)-(25 times 12)+ (5 times 6) -1=729$
So the number of "good" cuboïds is
$3375 - 729 = mathbf2646$.
answered Aug 17 at 13:32
Evargalo
869211
I think Gareth McCaughan has the right reasoning but a small calculation error. I'll copy his explanations here :
Suppose we leave the corners there. Then
Let's give coordinates to the cubes from 1 to 5 in length, width and height. A rectangular prism (= cuboïd) is defined by its smallest and biggest index (possibly equal) on each axis, so there are $(binom52+5)^3=15^3=3375$ of these.
How many of these are excluded by having no corners?
A cuboïd $[a,b]times[c,d]times[e,f]$ uses a corner iff ($a=1$ OR $b=5$) AND ($c=1$ OR $d=5$) AND ($e=1$ OR $f=5$). The number of ways to choose $(a,b)$ so that this happens is $9$, so there are $9^3=729$ of these.
Alternatively, by inclusion/exclusion:
There are $5 times 5 times 5=125$ cuboïds using the 'vanished' $(1,1,1)$ corner cube, and $8$ ways to pick a corner.
There are $5 times 5=25$ cuboïds using the 'vanished' $(1,1,1)$ and $(1,1,5)$ cubes, and $12$ ways to pick an edge.
There are $5$ cuboïds using the four 'vanished' $(1,1,1)$, $(1,5,1)$, $(1,5,5)$ and $(1,1,5)$ cubes on any given face, and $6$ ways to pick a face.
There is $1$ cuboïd using all eight 'vanished' corner cubes.
The number of 'wrong' cuboïds is then: $(125 times 8)-(25 times 12)+ (5 times 6) -1=729$
So the number of "good" cuboïds is
$3375 - 729 = mathbf2646$.
answered Aug 17 at 13:32
Evargalo
869211
edited Aug 17 at 21:04
answered Aug 17 at 13:32
Evargalo
869211
answered Aug 17 at 13:32
Evargalo
869211
answered Aug 17 at 13:32
Evargalo
869211
869211
Evargalo, I think you had it initially correct in your alternative explanation; your math no longer tallies in the final equation of your third paragraph.
– El-Guest
Aug 17 at 13:43
@El-Guest. Thanks, I've edited and I hope it is correct now.
– Evargalo
Aug 17 at 13:44
Looks much better. I can clean up some of the math formatting for you as well.
– El-Guest
Aug 17 at 13:46
1
Looks great! thanks again.
– Evargalo
Aug 17 at 13:49
2
This is slightly incorrect because it has "1 or 5" everywhere, where it should be "0 or 5". I confess myself slightly miffed that someone else is getting the coveted green checkmark for copying-and-pasting my answer, fixing a trivial error, and making another trivial error in turn, but never mind. I suppose there's the inclusion-exclusion bit. (I decided not to do it that way because the argument I actually gave was much easier to follow.)
– Gareth McCaughan♦
Aug 17 at 15:47
 |Â
show 2 more comments
Evargalo, I think you had it initially correct in your alternative explanation; your math no longer tallies in the final equation of your third paragraph.
– El-Guest
Aug 17 at 13:43
@El-Guest. Thanks, I've edited and I hope it is correct now.
– Evargalo
Aug 17 at 13:44
Looks much better. I can clean up some of the math formatting for you as well.
– El-Guest
Aug 17 at 13:46
1
Looks great! thanks again.
– Evargalo
Aug 17 at 13:49
2
This is slightly incorrect because it has "1 or 5" everywhere, where it should be "0 or 5". I confess myself slightly miffed that someone else is getting the coveted green checkmark for copying-and-pasting my answer, fixing a trivial error, and making another trivial error in turn, but never mind. I suppose there's the inclusion-exclusion bit. (I decided not to do it that way because the argument I actually gave was much easier to follow.)
– Gareth McCaughan♦
Aug 17 at 15:47
Evargalo, I think you had it initially correct in your alternative explanation; your math no longer tallies in the final equation of your third paragraph.
– El-Guest
Aug 17 at 13:43
Evargalo, I think you had it initially correct in your alternative explanation; your math no longer tallies in the final equation of your third paragraph.
– El-Guest
Aug 17 at 13:43
@El-Guest. Thanks, I've edited and I hope it is correct now.
– Evargalo
Aug 17 at 13:44
@El-Guest. Thanks, I've edited and I hope it is correct now.
– Evargalo
Aug 17 at 13:44
Looks much better. I can clean up some of the math formatting for you as well.
– El-Guest
Aug 17 at 13:46
Looks much better. I can clean up some of the math formatting for you as well.
– El-Guest
Aug 17 at 13:46
1
1
Looks great! thanks again.
– Evargalo
Aug 17 at 13:49
Looks great! thanks again.
– Evargalo
Aug 17 at 13:49
2
2
This is slightly incorrect because it has "1 or 5" everywhere, where it should be "0 or 5". I confess myself slightly miffed that someone else is getting the coveted green checkmark for copying-and-pasting my answer, fixing a trivial error, and making another trivial error in turn, but never mind. I suppose there's the inclusion-exclusion bit. (I decided not to do it that way because the argument I actually gave was much easier to follow.)
– Gareth McCaughan♦
Aug 17 at 15:47
This is slightly incorrect because it has "1 or 5" everywhere, where it should be "0 or 5". I confess myself slightly miffed that someone else is getting the coveted green checkmark for copying-and-pasting my answer, fixing a trivial error, and making another trivial error in turn, but never mind. I suppose there's the inclusion-exclusion bit. (I decided not to do it that way because the argument I actually gave was much easier to follow.)
– Gareth McCaughan♦
Aug 17 at 15:47
 |Â
show 2 more comments
up vote
3
down vote
A computer version in JavaScript:
<!doctype html>
<html>
<title>Cuboids</title>
<body>
<span id='out'></span><br>
<button onclick='go();'>go</button>
</body>
<script>
function go() c==14)) d++;
out.textContent=d;
d=0;
for (let a=0;a<5;a++)
for (let b=0;b<5;b++)
for (let c=0;c<5;c++)
d+=countCuboids(a,b,c);
out.textContent+=' :: '+d;
function countCuboids(a,b,c)
if (a%4==0 && b%4==0 && c%4==0) return 0;
if (a%4>0 && b%4==0 && c%4==0) return (4-a)*(5-b)*(5-c);
if (a%4==0 && b%4>0 && c%4==0) return (5-a)*(4-b)*(5-c);
if (a%4==0 && b%4==0 && c%4>0) return (5-a)*(5-b)*(4-c);
if (a%4>0 && b%4>0 && c%4==0) return (5-a)*(5-b)*(5-c)-(5-c);
if (a%4>0 && b%4==0 && c%4>0) return (5-a)*(5-b)*(5-c)-(5-b);
if (a%4==0 && b%4>0 && c%4>0) return (5-a)*(5-b)*(5-c)-(5-a);
if (a%4>0 && b%4>0 && c%4>0) return (5-a)*(5-b)*(5-c)-1;
</script>
</html>
The first version maps $0..14$ to $1,12,123,1234,12345,2,23,234,2345,3,34,345,4,45,5$ and counts anything that has a $1$ or a $5$ in all three loop variables.
Result: 729 (=$15^3-2646$)
The second uses an explicit function to calculate the number of right, up, back cuboids from a starting cube, and sums them.
Result: 2646
answered Aug 17 at 15:14
JonMark Perry
13.3k42666
add a comment |Â
up vote
3
down vote
A computer version in JavaScript:
<!doctype html>
<html>
<title>Cuboids</title>
<body>
<span id='out'></span><br>
<button onclick='go();'>go</button>
</body>
<script>
function go() c==14)) d++;
out.textContent=d;
d=0;
for (let a=0;a<5;a++)
for (let b=0;b<5;b++)
for (let c=0;c<5;c++)
d+=countCuboids(a,b,c);
out.textContent+=' :: '+d;
function countCuboids(a,b,c)
if (a%4==0 && b%4==0 && c%4==0) return 0;
if (a%4>0 && b%4==0 && c%4==0) return (4-a)*(5-b)*(5-c);
if (a%4==0 && b%4>0 && c%4==0) return (5-a)*(4-b)*(5-c);
if (a%4==0 && b%4==0 && c%4>0) return (5-a)*(5-b)*(4-c);
if (a%4>0 && b%4>0 && c%4==0) return (5-a)*(5-b)*(5-c)-(5-c);
if (a%4>0 && b%4==0 && c%4>0) return (5-a)*(5-b)*(5-c)-(5-b);
if (a%4==0 && b%4>0 && c%4>0) return (5-a)*(5-b)*(5-c)-(5-a);
if (a%4>0 && b%4>0 && c%4>0) return (5-a)*(5-b)*(5-c)-1;
</script>
</html>
The first version maps $0..14$ to $1,12,123,1234,12345,2,23,234,2345,3,34,345,4,45,5$ and counts anything that has a $1$ or a $5$ in all three loop variables.
Result: 729 (=$15^3-2646$)
The second uses an explicit function to calculate the number of right, up, back cuboids from a starting cube, and sums them.
Result: 2646
answered Aug 17 at 15:14
JonMark Perry
13.3k42666
add a comment |Â
up vote
3
down vote
up vote
3
down vote
A computer version in JavaScript:
<!doctype html>
<html>
<title>Cuboids</title>
<body>
<span id='out'></span><br>
<button onclick='go();'>go</button>
</body>
<script>
function go() c==14)) d++;
out.textContent=d;
d=0;
for (let a=0;a<5;a++)
for (let b=0;b<5;b++)
for (let c=0;c<5;c++)
d+=countCuboids(a,b,c);
out.textContent+=' :: '+d;
function countCuboids(a,b,c)
if (a%4==0 && b%4==0 && c%4==0) return 0;
if (a%4>0 && b%4==0 && c%4==0) return (4-a)*(5-b)*(5-c);
if (a%4==0 && b%4>0 && c%4==0) return (5-a)*(4-b)*(5-c);
if (a%4==0 && b%4==0 && c%4>0) return (5-a)*(5-b)*(4-c);
if (a%4>0 && b%4>0 && c%4==0) return (5-a)*(5-b)*(5-c)-(5-c);
if (a%4>0 && b%4==0 && c%4>0) return (5-a)*(5-b)*(5-c)-(5-b);
if (a%4==0 && b%4>0 && c%4>0) return (5-a)*(5-b)*(5-c)-(5-a);
if (a%4>0 && b%4>0 && c%4>0) return (5-a)*(5-b)*(5-c)-1;
</script>
</html>
The first version maps $0..14$ to $1,12,123,1234,12345,2,23,234,2345,3,34,345,4,45,5$ and counts anything that has a $1$ or a $5$ in all three loop variables.
Result: 729 (=$15^3-2646$)
The second uses an explicit function to calculate the number of right, up, back cuboids from a starting cube, and sums them.
Result: 2646
answered Aug 17 at 15:14
JonMark Perry
13.3k42666
A computer version in JavaScript:
<!doctype html>
<html>
<title>Cuboids</title>
<body>
<span id='out'></span><br>
<button onclick='go();'>go</button>
</body>
<script>
function go() c==14)) d++;
out.textContent=d;
d=0;
for (let a=0;a<5;a++)
for (let b=0;b<5;b++)
for (let c=0;c<5;c++)
d+=countCuboids(a,b,c);
out.textContent+=' :: '+d;
function countCuboids(a,b,c)
if (a%4==0 && b%4==0 && c%4==0) return 0;
if (a%4>0 && b%4==0 && c%4==0) return (4-a)*(5-b)*(5-c);
if (a%4==0 && b%4>0 && c%4==0) return (5-a)*(4-b)*(5-c);
if (a%4==0 && b%4==0 && c%4>0) return (5-a)*(5-b)*(4-c);
if (a%4>0 && b%4>0 && c%4==0) return (5-a)*(5-b)*(5-c)-(5-c);
if (a%4>0 && b%4==0 && c%4>0) return (5-a)*(5-b)*(5-c)-(5-b);
if (a%4==0 && b%4>0 && c%4>0) return (5-a)*(5-b)*(5-c)-(5-a);
if (a%4>0 && b%4>0 && c%4>0) return (5-a)*(5-b)*(5-c)-1;
</script>
</html>
The first version maps $0..14$ to $1,12,123,1234,12345,2,23,234,2345,3,34,345,4,45,5$ and counts anything that has a $1$ or a $5$ in all three loop variables.
Result: 729 (=$15^3-2646$)
The second uses an explicit function to calculate the number of right, up, back cuboids from a starting cube, and sums them.
Result: 2646
answered Aug 17 at 15:14
JonMark Perry
13.3k42666
answered Aug 17 at 15:14
JonMark Perry
13.3k42666
answered Aug 17 at 15:14
JonMark Perry
13.3k42666
answered Aug 17 at 15:14
JonMark Perry
13.3k42666
13.3k42666
add a comment |Â
add a comment |Â
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Does a cube count as a rectangular prism??
– Kevin L
Aug 17 at 12:48
1
we dont know if any group of prism creates a cube and since dimensions are not given, yes you will count cubes as prisms.
– Oray
Aug 17 at 12:50
ok gonna start counting :)
– Kevin L
Aug 17 at 12:51
and according to wyzant.com/resources/answers/4393/… cube is also rectangular prism.
– Oray
Aug 17 at 12:52
Since Gareth McCaugham has corrected his slight mistake you should credit him with the accepted answer, for anteriority and for the redaction effort that I just copied and pasted.
– Evargalo
Aug 17 at 20:57