How many integer solutions are there to the inequality $x_1 + x_2 + x_3 leq 17$ if we require that $x_1 geq 1$, $x_2 geq 2$, $x_3 geq 3$?
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How many integer solutions are there to the inequality
$x_1 + x_2 + x_3 leq 17$,
if we require that
$x_1 geq 1$, $x_2 geq 2$, $x_3 geq 3$?
The way I did it:
$x_4 = 17-x_1-x_2-x_3-x_4$
$x_1+x_2+x_3+x_4 = 17$
$x_1= x_i + 1$
$x_2= x_ii+2$
$x_3= x_iii+3$
$x_4= x_iiii$
So,
$x_i+1+x_ii+2+x_iii+3+x_iiii= 17$
$x_i+x_ii+x_iii+x_iiii=11$
$3$ bars, $binom143 =364$ different ways?
Is this the right approach? correct answer?
combinatorics combinations
edited Jul 5 at 1:21
N. F. Taussig
39.4k93153
asked Jul 5 at 0:37
Sacha E
416
add a comment |Â
up vote
0
down vote
favorite
How many integer solutions are there to the inequality
$x_1 + x_2 + x_3 leq 17$,
if we require that
$x_1 geq 1$, $x_2 geq 2$, $x_3 geq 3$?
The way I did it:
$x_4 = 17-x_1-x_2-x_3-x_4$
$x_1+x_2+x_3+x_4 = 17$
$x_1= x_i + 1$
$x_2= x_ii+2$
$x_3= x_iii+3$
$x_4= x_iiii$
So,
$x_i+1+x_ii+2+x_iii+3+x_iiii= 17$
$x_i+x_ii+x_iii+x_iiii=11$
$3$ bars, $binom143 =364$ different ways?
Is this the right approach? correct answer?
combinatorics combinations
edited Jul 5 at 1:21
N. F. Taussig
39.4k93153
asked Jul 5 at 0:37
Sacha E
416
Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site. You have solved the problem correctly.
– N. F. Taussig
Jul 5 at 1:22
If you need only the solution - brute-force counting works just fine, as the numbers are small. I got 364 solutions.
– fanvacoolt
Jul 5 at 2:49
Try summation over generating function.
– Nong
Jul 5 at 4:27
1
Sacha E, do you know about accepting answers you find to be helpful? If you find one post particularly helpful, per post, you can click on the greyed out $checkmark$, and it will turn green. You can accept one answer per post, and when you do, the answerer gets extra points, and you yourself earn two points for every accepted answer.
– amWhy
Jul 6 at 23:01
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How many integer solutions are there to the inequality
$x_1 + x_2 + x_3 leq 17$,
if we require that
$x_1 geq 1$, $x_2 geq 2$, $x_3 geq 3$?
The way I did it:
$x_4 = 17-x_1-x_2-x_3-x_4$
$x_1+x_2+x_3+x_4 = 17$
$x_1= x_i + 1$
$x_2= x_ii+2$
$x_3= x_iii+3$
$x_4= x_iiii$
So,
$x_i+1+x_ii+2+x_iii+3+x_iiii= 17$
$x_i+x_ii+x_iii+x_iiii=11$
$3$ bars, $binom143 =364$ different ways?
Is this the right approach? correct answer?
combinatorics combinations
edited Jul 5 at 1:21
N. F. Taussig
39.4k93153
asked Jul 5 at 0:37
Sacha E
416
How many integer solutions are there to the inequality
$x_1 + x_2 + x_3 leq 17$,
if we require that
$x_1 geq 1$, $x_2 geq 2$, $x_3 geq 3$?
The way I did it:
$x_4 = 17-x_1-x_2-x_3-x_4$
$x_1+x_2+x_3+x_4 = 17$
$x_1= x_i + 1$
$x_2= x_ii+2$
$x_3= x_iii+3$
$x_4= x_iiii$
So,
$x_i+1+x_ii+2+x_iii+3+x_iiii= 17$
$x_i+x_ii+x_iii+x_iiii=11$
$3$ bars, $binom143 =364$ different ways?
Is this the right approach? correct answer?
combinatorics combinations
combinatorics combinations
edited Jul 5 at 1:21
N. F. Taussig
39.4k93153
asked Jul 5 at 0:37
Sacha E
416
edited Jul 5 at 1:21
N. F. Taussig
39.4k93153
asked Jul 5 at 0:37
Sacha E
416
edited Jul 5 at 1:21
N. F. Taussig
39.4k93153
edited Jul 5 at 1:21
N. F. Taussig
39.4k93153
edited Jul 5 at 1:21
N. F. Taussig
39.4k93153
39.4k93153
asked Jul 5 at 0:37
Sacha E
416
asked Jul 5 at 0:37
Sacha E
416
asked Jul 5 at 0:37
Sacha E
416
416
Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site. You have solved the problem correctly.
– N. F. Taussig
Jul 5 at 1:22
If you need only the solution - brute-force counting works just fine, as the numbers are small. I got 364 solutions.
– fanvacoolt
Jul 5 at 2:49
Try summation over generating function.
– Nong
Jul 5 at 4:27
1
Sacha E, do you know about accepting answers you find to be helpful? If you find one post particularly helpful, per post, you can click on the greyed out $checkmark$, and it will turn green. You can accept one answer per post, and when you do, the answerer gets extra points, and you yourself earn two points for every accepted answer.
– amWhy
Jul 6 at 23:01
add a comment |Â
Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site. You have solved the problem correctly.
– N. F. Taussig
Jul 5 at 1:22
If you need only the solution - brute-force counting works just fine, as the numbers are small. I got 364 solutions.
– fanvacoolt
Jul 5 at 2:49
Try summation over generating function.
– Nong
Jul 5 at 4:27
1
Sacha E, do you know about accepting answers you find to be helpful? If you find one post particularly helpful, per post, you can click on the greyed out $checkmark$, and it will turn green. You can accept one answer per post, and when you do, the answerer gets extra points, and you yourself earn two points for every accepted answer.
– amWhy
Jul 6 at 23:01
Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site. You have solved the problem correctly.
– N. F. Taussig
Jul 5 at 1:22
Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site. You have solved the problem correctly.
– N. F. Taussig
Jul 5 at 1:22
If you need only the solution - brute-force counting works just fine, as the numbers are small. I got 364 solutions.
– fanvacoolt
Jul 5 at 2:49
If you need only the solution - brute-force counting works just fine, as the numbers are small. I got 364 solutions.
– fanvacoolt
Jul 5 at 2:49
Try summation over generating function.
– Nong
Jul 5 at 4:27
Try summation over generating function.
– Nong
Jul 5 at 4:27
1
1
Sacha E, do you know about accepting answers you find to be helpful? If you find one post particularly helpful, per post, you can click on the greyed out $checkmark$, and it will turn green. You can accept one answer per post, and when you do, the answerer gets extra points, and you yourself earn two points for every accepted answer.
– amWhy
Jul 6 at 23:01
Sacha E, do you know about accepting answers you find to be helpful? If you find one post particularly helpful, per post, you can click on the greyed out $checkmark$, and it will turn green. You can accept one answer per post, and when you do, the answerer gets extra points, and you yourself earn two points for every accepted answer.
– amWhy
Jul 6 at 23:01
add a comment |Â
1 Answer
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for the inequality x1+x2+x3<=17,we may introduce a fourth variable say x4,so as to solve the number of integer solutions to;x1+x2+x3+x4=17 with x1>=1,x2>=2 and x3>=3.However, the three constraints implies x1-1>=o,x2-2>=o and x3-3>=o.so if we let x1-1=a,x2-2=b,x3-3=c, x4=d with x4>=o, we therefore solve the number of integer solutions to (a+1)+(b+2)+(c+3)+d=17 which implies a+b+c+d=11 with a,b,c,d>=o which is (11+4-1)Combination(4-1)=(14)combination(3)=364.thanks.
answered Aug 31 at 23:52
Ekpe Owai
1
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-1
down vote
for the inequality x1+x2+x3<=17,we may introduce a fourth variable say x4,so as to solve the number of integer solutions to;x1+x2+x3+x4=17 with x1>=1,x2>=2 and x3>=3.However, the three constraints implies x1-1>=o,x2-2>=o and x3-3>=o.so if we let x1-1=a,x2-2=b,x3-3=c, x4=d with x4>=o, we therefore solve the number of integer solutions to (a+1)+(b+2)+(c+3)+d=17 which implies a+b+c+d=11 with a,b,c,d>=o which is (11+4-1)Combination(4-1)=(14)combination(3)=364.thanks.
answered Aug 31 at 23:52
Ekpe Owai
1
add a comment |Â
up vote
-1
down vote
for the inequality x1+x2+x3<=17,we may introduce a fourth variable say x4,so as to solve the number of integer solutions to;x1+x2+x3+x4=17 with x1>=1,x2>=2 and x3>=3.However, the three constraints implies x1-1>=o,x2-2>=o and x3-3>=o.so if we let x1-1=a,x2-2=b,x3-3=c, x4=d with x4>=o, we therefore solve the number of integer solutions to (a+1)+(b+2)+(c+3)+d=17 which implies a+b+c+d=11 with a,b,c,d>=o which is (11+4-1)Combination(4-1)=(14)combination(3)=364.thanks.
answered Aug 31 at 23:52
Ekpe Owai
1
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
for the inequality x1+x2+x3<=17,we may introduce a fourth variable say x4,so as to solve the number of integer solutions to;x1+x2+x3+x4=17 with x1>=1,x2>=2 and x3>=3.However, the three constraints implies x1-1>=o,x2-2>=o and x3-3>=o.so if we let x1-1=a,x2-2=b,x3-3=c, x4=d with x4>=o, we therefore solve the number of integer solutions to (a+1)+(b+2)+(c+3)+d=17 which implies a+b+c+d=11 with a,b,c,d>=o which is (11+4-1)Combination(4-1)=(14)combination(3)=364.thanks.
answered Aug 31 at 23:52
Ekpe Owai
1
for the inequality x1+x2+x3<=17,we may introduce a fourth variable say x4,so as to solve the number of integer solutions to;x1+x2+x3+x4=17 with x1>=1,x2>=2 and x3>=3.However, the three constraints implies x1-1>=o,x2-2>=o and x3-3>=o.so if we let x1-1=a,x2-2=b,x3-3=c, x4=d with x4>=o, we therefore solve the number of integer solutions to (a+1)+(b+2)+(c+3)+d=17 which implies a+b+c+d=11 with a,b,c,d>=o which is (11+4-1)Combination(4-1)=(14)combination(3)=364.thanks.
answered Aug 31 at 23:52
Ekpe Owai
1
answered Aug 31 at 23:52
Ekpe Owai
1
answered Aug 31 at 23:52
Ekpe Owai
1
answered Aug 31 at 23:52
Ekpe Owai
1
1
add a comment |Â
add a comment |Â
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Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site. You have solved the problem correctly.
– N. F. Taussig
Jul 5 at 1:22
If you need only the solution - brute-force counting works just fine, as the numbers are small. I got 364 solutions.
– fanvacoolt
Jul 5 at 2:49
Try summation over generating function.
– Nong
Jul 5 at 4:27
1
Sacha E, do you know about accepting answers you find to be helpful? If you find one post particularly helpful, per post, you can click on the greyed out $checkmark$, and it will turn green. You can accept one answer per post, and when you do, the answerer gets extra points, and you yourself earn two points for every accepted answer.
– amWhy
Jul 6 at 23:01