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number of possibilities for a 6 digit number with contraints
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We have a 6 digit number _ _ _ _ _ _
the constraints are :
1. the 6th digit is the sum of 2th and the 4th digit.
2. the 5th digit is the sum of 1th and the 3th digit.
3. digits can be from 1 to 9. Repetition of digits is allowed.
how many possibilities are there?
I have tried something, written down every possibly combination of sum of two digits that doesn't exceed 9. I got 36. then multiplied it by 4. but it's not correct
combinatorics combinations
asked Jul 30 at 13:01
BinaryVeil
1042
 |Â
show 2 more comments
up vote
0
down vote
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We have a 6 digit number _ _ _ _ _ _
the constraints are :
1. the 6th digit is the sum of 2th and the 4th digit.
2. the 5th digit is the sum of 1th and the 3th digit.
3. digits can be from 1 to 9. Repetition of digits is allowed.
how many possibilities are there?
I have tried something, written down every possibly combination of sum of two digits that doesn't exceed 9. I got 36. then multiplied it by 4. but it's not correct
combinatorics combinations
asked Jul 30 at 13:01
BinaryVeil
1042
Surely you have tried something, no? When you edit your post to include your efforts, I also suggest that you indicate which digit you mean by (say) the $6^th$...if you mean the leading digit then, presumably, you want to exclude the case where it is $0$.
– lulu
Jul 30 at 13:09
I've edited the post. the digits can be from 1 to 9. I have tried something, written down every possibly combination of sum of two digits that doesn't exceed 9. I got 36. then multiplied it by 4. but it's not correct
– BinaryVeil
Jul 30 at 13:21
Let's count the unordered pairs of non-zero digits that add to a digit: if the max is $9$ then there are none. If the max is $8$ then there is one. If $7$ then two. If $6$ then three. If $5$ then four. if $4$ then four. if $3$ then three. if $2$ then two. if $1$ then one. Thus $1+2+3+4+4+3+2+1=20$. Of these there are four doubles and $16$ non-doubles. Can you finish from here? Why would you multiply by $4$?
– lulu
Jul 30 at 13:27
thank you. Unfortunately I didn't quite get it.. The options for digits are :1,1 ; 1,2; 1,3; 1,4; 1,5; 1,6; 1,7; 1,8; 2,1 ; 2,2; 2,3 ; 2,4; 2,5 ; 2,6 ; 2,7; 3,1 ; 3,2; 3,3; 3,4 ; 3,5; 3,6; 4,1; 4,2; 4,3 ; 4,4; 4,5; 5,1; 5,2; 5,3; 5;4 6,1; 6,2 ; 6,3 7,1 ; 7,2 8;1
– BinaryVeil
Jul 30 at 13:32
2
I am also getting $36$ ordered pairs, four doubles and sixteen non-doubles among the unordered pairs gives $4+2times 16=4+32=36$. But I can't understand why you would multiply by $4$? you need to pick two ordered pairs. That's $36^2$.
– lulu
Jul 30 at 13:33
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We have a 6 digit number _ _ _ _ _ _
the constraints are :
1. the 6th digit is the sum of 2th and the 4th digit.
2. the 5th digit is the sum of 1th and the 3th digit.
3. digits can be from 1 to 9. Repetition of digits is allowed.
how many possibilities are there?
I have tried something, written down every possibly combination of sum of two digits that doesn't exceed 9. I got 36. then multiplied it by 4. but it's not correct
combinatorics combinations
asked Jul 30 at 13:01
BinaryVeil
1042
We have a 6 digit number _ _ _ _ _ _
the constraints are :
1. the 6th digit is the sum of 2th and the 4th digit.
2. the 5th digit is the sum of 1th and the 3th digit.
3. digits can be from 1 to 9. Repetition of digits is allowed.
how many possibilities are there?
I have tried something, written down every possibly combination of sum of two digits that doesn't exceed 9. I got 36. then multiplied it by 4. but it's not correct
combinatorics combinations
asked Jul 30 at 13:01
BinaryVeil
1042
edited Jul 30 at 13:23
asked Jul 30 at 13:01
BinaryVeil
1042
asked Jul 30 at 13:01
BinaryVeil
1042
asked Jul 30 at 13:01
BinaryVeil
1042
1042
Surely you have tried something, no? When you edit your post to include your efforts, I also suggest that you indicate which digit you mean by (say) the $6^th$...if you mean the leading digit then, presumably, you want to exclude the case where it is $0$.
– lulu
Jul 30 at 13:09
I've edited the post. the digits can be from 1 to 9. I have tried something, written down every possibly combination of sum of two digits that doesn't exceed 9. I got 36. then multiplied it by 4. but it's not correct
– BinaryVeil
Jul 30 at 13:21
Let's count the unordered pairs of non-zero digits that add to a digit: if the max is $9$ then there are none. If the max is $8$ then there is one. If $7$ then two. If $6$ then three. If $5$ then four. if $4$ then four. if $3$ then three. if $2$ then two. if $1$ then one. Thus $1+2+3+4+4+3+2+1=20$. Of these there are four doubles and $16$ non-doubles. Can you finish from here? Why would you multiply by $4$?
– lulu
Jul 30 at 13:27
thank you. Unfortunately I didn't quite get it.. The options for digits are :1,1 ; 1,2; 1,3; 1,4; 1,5; 1,6; 1,7; 1,8; 2,1 ; 2,2; 2,3 ; 2,4; 2,5 ; 2,6 ; 2,7; 3,1 ; 3,2; 3,3; 3,4 ; 3,5; 3,6; 4,1; 4,2; 4,3 ; 4,4; 4,5; 5,1; 5,2; 5,3; 5;4 6,1; 6,2 ; 6,3 7,1 ; 7,2 8;1
– BinaryVeil
Jul 30 at 13:32
2
I am also getting $36$ ordered pairs, four doubles and sixteen non-doubles among the unordered pairs gives $4+2times 16=4+32=36$. But I can't understand why you would multiply by $4$? you need to pick two ordered pairs. That's $36^2$.
– lulu
Jul 30 at 13:33
 |Â
show 2 more comments
Surely you have tried something, no? When you edit your post to include your efforts, I also suggest that you indicate which digit you mean by (say) the $6^th$...if you mean the leading digit then, presumably, you want to exclude the case where it is $0$.
– lulu
Jul 30 at 13:09
I've edited the post. the digits can be from 1 to 9. I have tried something, written down every possibly combination of sum of two digits that doesn't exceed 9. I got 36. then multiplied it by 4. but it's not correct
– BinaryVeil
Jul 30 at 13:21
Let's count the unordered pairs of non-zero digits that add to a digit: if the max is $9$ then there are none. If the max is $8$ then there is one. If $7$ then two. If $6$ then three. If $5$ then four. if $4$ then four. if $3$ then three. if $2$ then two. if $1$ then one. Thus $1+2+3+4+4+3+2+1=20$. Of these there are four doubles and $16$ non-doubles. Can you finish from here? Why would you multiply by $4$?
– lulu
Jul 30 at 13:27
thank you. Unfortunately I didn't quite get it.. The options for digits are :1,1 ; 1,2; 1,3; 1,4; 1,5; 1,6; 1,7; 1,8; 2,1 ; 2,2; 2,3 ; 2,4; 2,5 ; 2,6 ; 2,7; 3,1 ; 3,2; 3,3; 3,4 ; 3,5; 3,6; 4,1; 4,2; 4,3 ; 4,4; 4,5; 5,1; 5,2; 5,3; 5;4 6,1; 6,2 ; 6,3 7,1 ; 7,2 8;1
– BinaryVeil
Jul 30 at 13:32
2
I am also getting $36$ ordered pairs, four doubles and sixteen non-doubles among the unordered pairs gives $4+2times 16=4+32=36$. But I can't understand why you would multiply by $4$? you need to pick two ordered pairs. That's $36^2$.
– lulu
Jul 30 at 13:33
Surely you have tried something, no? When you edit your post to include your efforts, I also suggest that you indicate which digit you mean by (say) the $6^th$...if you mean the leading digit then, presumably, you want to exclude the case where it is $0$.
– lulu
Jul 30 at 13:09
Surely you have tried something, no? When you edit your post to include your efforts, I also suggest that you indicate which digit you mean by (say) the $6^th$...if you mean the leading digit then, presumably, you want to exclude the case where it is $0$.
– lulu
Jul 30 at 13:09
I've edited the post. the digits can be from 1 to 9. I have tried something, written down every possibly combination of sum of two digits that doesn't exceed 9. I got 36. then multiplied it by 4. but it's not correct
– BinaryVeil
Jul 30 at 13:21
I've edited the post. the digits can be from 1 to 9. I have tried something, written down every possibly combination of sum of two digits that doesn't exceed 9. I got 36. then multiplied it by 4. but it's not correct
– BinaryVeil
Jul 30 at 13:21
Let's count the unordered pairs of non-zero digits that add to a digit: if the max is $9$ then there are none. If the max is $8$ then there is one. If $7$ then two. If $6$ then three. If $5$ then four. if $4$ then four. if $3$ then three. if $2$ then two. if $1$ then one. Thus $1+2+3+4+4+3+2+1=20$. Of these there are four doubles and $16$ non-doubles. Can you finish from here? Why would you multiply by $4$?
– lulu
Jul 30 at 13:27
Let's count the unordered pairs of non-zero digits that add to a digit: if the max is $9$ then there are none. If the max is $8$ then there is one. If $7$ then two. If $6$ then three. If $5$ then four. if $4$ then four. if $3$ then three. if $2$ then two. if $1$ then one. Thus $1+2+3+4+4+3+2+1=20$. Of these there are four doubles and $16$ non-doubles. Can you finish from here? Why would you multiply by $4$?
– lulu
Jul 30 at 13:27
thank you. Unfortunately I didn't quite get it.. The options for digits are :
1,1 ; 1,2; 1,3; 1,4; 1,5; 1,6; 1,7; 1,8; 2,1 ; 2,2; 2,3 ; 2,4; 2,5 ; 2,6 ; 2,7; 3,1 ; 3,2; 3,3; 3,4 ; 3,5; 3,6; 4,1; 4,2; 4,3 ; 4,4; 4,5; 5,1; 5,2; 5,3; 5;4 6,1; 6,2 ; 6,3 7,1 ; 7,2 8;1 – BinaryVeil
Jul 30 at 13:32
thank you. Unfortunately I didn't quite get it.. The options for digits are :
1,1 ; 1,2; 1,3; 1,4; 1,5; 1,6; 1,7; 1,8; 2,1 ; 2,2; 2,3 ; 2,4; 2,5 ; 2,6 ; 2,7; 3,1 ; 3,2; 3,3; 3,4 ; 3,5; 3,6; 4,1; 4,2; 4,3 ; 4,4; 4,5; 5,1; 5,2; 5,3; 5;4 6,1; 6,2 ; 6,3 7,1 ; 7,2 8;1 – BinaryVeil
Jul 30 at 13:32
2
2
I am also getting $36$ ordered pairs, four doubles and sixteen non-doubles among the unordered pairs gives $4+2times 16=4+32=36$. But I can't understand why you would multiply by $4$? you need to pick two ordered pairs. That's $36^2$.
– lulu
Jul 30 at 13:33
I am also getting $36$ ordered pairs, four doubles and sixteen non-doubles among the unordered pairs gives $4+2times 16=4+32=36$. But I can't understand why you would multiply by $4$? you need to pick two ordered pairs. That's $36^2$.
– lulu
Jul 30 at 13:33
 |Â
show 2 more comments
1 Answer
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Let's call the digits $d_1, d_2, ldots d_6$. Clearly the first four digits determine the whole number, but there are constraints on those digits in order to ensure that $d_5$ and $d_6$ stay in range.
So, for example, choosing $d_1=1$ gives $8$ options for $d_3$ to keep $d_5$ in range. Choosing $d_1=2$ gives $7$ options for $d_3$, etc., leading to a total of $8+7+6+cdots+1=36$ options for the $d_1,d_3$ set. This agrees with part of your reasoning.
Independently of that choice, by the same calculations we similarly have $36$ options for $d_2,d_4$. Clearly any one choice for $d_1,d_3$ can be associated with any choice for $d_2,d_4,$ so overall the total choices for the whole six digit number are $36cdot 36 = 1296$.
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
accepted
Let's call the digits $d_1, d_2, ldots d_6$. Clearly the first four digits determine the whole number, but there are constraints on those digits in order to ensure that $d_5$ and $d_6$ stay in range.
So, for example, choosing $d_1=1$ gives $8$ options for $d_3$ to keep $d_5$ in range. Choosing $d_1=2$ gives $7$ options for $d_3$, etc., leading to a total of $8+7+6+cdots+1=36$ options for the $d_1,d_3$ set. This agrees with part of your reasoning.
Independently of that choice, by the same calculations we similarly have $36$ options for $d_2,d_4$. Clearly any one choice for $d_1,d_3$ can be associated with any choice for $d_2,d_4,$ so overall the total choices for the whole six digit number are $36cdot 36 = 1296$.
add a comment |Â
up vote
1
down vote
accepted
Let's call the digits $d_1, d_2, ldots d_6$. Clearly the first four digits determine the whole number, but there are constraints on those digits in order to ensure that $d_5$ and $d_6$ stay in range.
So, for example, choosing $d_1=1$ gives $8$ options for $d_3$ to keep $d_5$ in range. Choosing $d_1=2$ gives $7$ options for $d_3$, etc., leading to a total of $8+7+6+cdots+1=36$ options for the $d_1,d_3$ set. This agrees with part of your reasoning.
Independently of that choice, by the same calculations we similarly have $36$ options for $d_2,d_4$. Clearly any one choice for $d_1,d_3$ can be associated with any choice for $d_2,d_4,$ so overall the total choices for the whole six digit number are $36cdot 36 = 1296$.
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let's call the digits $d_1, d_2, ldots d_6$. Clearly the first four digits determine the whole number, but there are constraints on those digits in order to ensure that $d_5$ and $d_6$ stay in range.
So, for example, choosing $d_1=1$ gives $8$ options for $d_3$ to keep $d_5$ in range. Choosing $d_1=2$ gives $7$ options for $d_3$, etc., leading to a total of $8+7+6+cdots+1=36$ options for the $d_1,d_3$ set. This agrees with part of your reasoning.
Independently of that choice, by the same calculations we similarly have $36$ options for $d_2,d_4$. Clearly any one choice for $d_1,d_3$ can be associated with any choice for $d_2,d_4,$ so overall the total choices for the whole six digit number are $36cdot 36 = 1296$.
Let's call the digits $d_1, d_2, ldots d_6$. Clearly the first four digits determine the whole number, but there are constraints on those digits in order to ensure that $d_5$ and $d_6$ stay in range.
So, for example, choosing $d_1=1$ gives $8$ options for $d_3$ to keep $d_5$ in range. Choosing $d_1=2$ gives $7$ options for $d_3$, etc., leading to a total of $8+7+6+cdots+1=36$ options for the $d_1,d_3$ set. This agrees with part of your reasoning.
Independently of that choice, by the same calculations we similarly have $36$ options for $d_2,d_4$. Clearly any one choice for $d_1,d_3$ can be associated with any choice for $d_2,d_4,$ so overall the total choices for the whole six digit number are $36cdot 36 = 1296$.
answered Jul 30 at 16:44
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Joffan
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Surely you have tried something, no? When you edit your post to include your efforts, I also suggest that you indicate which digit you mean by (say) the $6^th$...if you mean the leading digit then, presumably, you want to exclude the case where it is $0$.
– lulu
Jul 30 at 13:09
I've edited the post. the digits can be from 1 to 9. I have tried something, written down every possibly combination of sum of two digits that doesn't exceed 9. I got 36. then multiplied it by 4. but it's not correct
– BinaryVeil
Jul 30 at 13:21
Let's count the unordered pairs of non-zero digits that add to a digit: if the max is $9$ then there are none. If the max is $8$ then there is one. If $7$ then two. If $6$ then three. If $5$ then four. if $4$ then four. if $3$ then three. if $2$ then two. if $1$ then one. Thus $1+2+3+4+4+3+2+1=20$. Of these there are four doubles and $16$ non-doubles. Can you finish from here? Why would you multiply by $4$?
– lulu
Jul 30 at 13:27
thank you. Unfortunately I didn't quite get it.. The options for digits are :
1,1 ; 1,2; 1,3; 1,4; 1,5; 1,6; 1,7; 1,8; 2,1 ; 2,2; 2,3 ; 2,4; 2,5 ; 2,6 ; 2,7; 3,1 ; 3,2; 3,3; 3,4 ; 3,5; 3,6; 4,1; 4,2; 4,3 ; 4,4; 4,5; 5,1; 5,2; 5,3; 5;4 6,1; 6,2 ; 6,3 7,1 ; 7,2 8;1– BinaryVeil
Jul 30 at 13:32
2
I am also getting $36$ ordered pairs, four doubles and sixteen non-doubles among the unordered pairs gives $4+2times 16=4+32=36$. But I can't understand why you would multiply by $4$? you need to pick two ordered pairs. That's $36^2$.
– lulu
Jul 30 at 13:33