Possible number of sequences such that $x_i = 1$ or $ 2$ and $sum_1^n x_i = 10$
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How many finite sequences are there such that $x_i = 1$ or $ 2$ and $sum_1^n x_i = 10$ $?$
Now I did it this way:
number of $1$'s $ $:$ $ $0$ ,$2$ ,$4$ ,$6$ , $8$ , $10$ and corresponding
number of $2$'s $ $:$ $ $5$ ,$4$ ,$3$ ,$2$ , $1$ , $0$
So the number such sequence is $1+ binom64 + binom73 +binom82 +binom91 + 1$ = $124$
But the answer says it is $89?$
Where is the mistake$?$
combinatorics permutations
edited Aug 10 at 16:23
Cloud JR
528415
asked Aug 2 '15 at 18:36
user118494
2,66911135
add a comment |Â
up vote
0
down vote
favorite
How many finite sequences are there such that $x_i = 1$ or $ 2$ and $sum_1^n x_i = 10$ $?$
Now I did it this way:
number of $1$'s $ $:$ $ $0$ ,$2$ ,$4$ ,$6$ , $8$ , $10$ and corresponding
number of $2$'s $ $:$ $ $5$ ,$4$ ,$3$ ,$2$ , $1$ , $0$
So the number such sequence is $1+ binom64 + binom73 +binom82 +binom91 + 1$ = $124$
But the answer says it is $89?$
Where is the mistake$?$
combinatorics permutations
edited Aug 10 at 16:23
Cloud JR
528415
asked Aug 2 '15 at 18:36
user118494
2,66911135
Check your sum again. I got $89$ using the same method you used.
– lulu
Aug 2 '15 at 18:42
Oops! That was so silly,posting it here. Answer did not match and I thought formula was wrong.
– user118494
Aug 2 '15 at 18:45
No worries. Combinatorics are simply confusing.
– lulu
Aug 2 '15 at 18:49
2
By the way, if $a_k$ is the number of such sequences of length $k$, one can show quite easily (no binomial coefficients) that $a_n=a_n-1+a_n-2$. It is easy to find $a_1$ and $a_2$. Now use the recurrence to work your way up to $10$. We get the familiar Fibonacci numbers.
– André Nicolas
Aug 2 '15 at 18:56
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How many finite sequences are there such that $x_i = 1$ or $ 2$ and $sum_1^n x_i = 10$ $?$
Now I did it this way:
number of $1$'s $ $:$ $ $0$ ,$2$ ,$4$ ,$6$ , $8$ , $10$ and corresponding
number of $2$'s $ $:$ $ $5$ ,$4$ ,$3$ ,$2$ , $1$ , $0$
So the number such sequence is $1+ binom64 + binom73 +binom82 +binom91 + 1$ = $124$
But the answer says it is $89?$
Where is the mistake$?$
combinatorics permutations
edited Aug 10 at 16:23
Cloud JR
528415
asked Aug 2 '15 at 18:36
user118494
2,66911135
How many finite sequences are there such that $x_i = 1$ or $ 2$ and $sum_1^n x_i = 10$ $?$
Now I did it this way:
number of $1$'s $ $:$ $ $0$ ,$2$ ,$4$ ,$6$ , $8$ , $10$ and corresponding
number of $2$'s $ $:$ $ $5$ ,$4$ ,$3$ ,$2$ , $1$ , $0$
So the number such sequence is $1+ binom64 + binom73 +binom82 +binom91 + 1$ = $124$
But the answer says it is $89?$
Where is the mistake$?$
combinatorics permutations
edited Aug 10 at 16:23
Cloud JR
528415
asked Aug 2 '15 at 18:36
user118494
2,66911135
edited Aug 10 at 16:23
Cloud JR
528415
edited Aug 10 at 16:23
Cloud JR
528415
edited Aug 10 at 16:23
Cloud JR
528415
528415
asked Aug 2 '15 at 18:36
user118494
2,66911135
asked Aug 2 '15 at 18:36
user118494
2,66911135
asked Aug 2 '15 at 18:36
user118494
2,66911135
2,66911135
Check your sum again. I got $89$ using the same method you used.
– lulu
Aug 2 '15 at 18:42
Oops! That was so silly,posting it here. Answer did not match and I thought formula was wrong.
– user118494
Aug 2 '15 at 18:45
No worries. Combinatorics are simply confusing.
– lulu
Aug 2 '15 at 18:49
2
By the way, if $a_k$ is the number of such sequences of length $k$, one can show quite easily (no binomial coefficients) that $a_n=a_n-1+a_n-2$. It is easy to find $a_1$ and $a_2$. Now use the recurrence to work your way up to $10$. We get the familiar Fibonacci numbers.
– André Nicolas
Aug 2 '15 at 18:56
add a comment |Â
Check your sum again. I got $89$ using the same method you used.
– lulu
Aug 2 '15 at 18:42
Oops! That was so silly,posting it here. Answer did not match and I thought formula was wrong.
– user118494
Aug 2 '15 at 18:45
No worries. Combinatorics are simply confusing.
– lulu
Aug 2 '15 at 18:49
2
By the way, if $a_k$ is the number of such sequences of length $k$, one can show quite easily (no binomial coefficients) that $a_n=a_n-1+a_n-2$. It is easy to find $a_1$ and $a_2$. Now use the recurrence to work your way up to $10$. We get the familiar Fibonacci numbers.
– André Nicolas
Aug 2 '15 at 18:56
Check your sum again. I got $89$ using the same method you used.
– lulu
Aug 2 '15 at 18:42
Check your sum again. I got $89$ using the same method you used.
– lulu
Aug 2 '15 at 18:42
Oops! That was so silly,posting it here. Answer did not match and I thought formula was wrong.
– user118494
Aug 2 '15 at 18:45
Oops! That was so silly,posting it here. Answer did not match and I thought formula was wrong.
– user118494
Aug 2 '15 at 18:45
No worries. Combinatorics are simply confusing.
– lulu
Aug 2 '15 at 18:49
No worries. Combinatorics are simply confusing.
– lulu
Aug 2 '15 at 18:49
2
2
By the way, if $a_k$ is the number of such sequences of length $k$, one can show quite easily (no binomial coefficients) that $a_n=a_n-1+a_n-2$. It is easy to find $a_1$ and $a_2$. Now use the recurrence to work your way up to $10$. We get the familiar Fibonacci numbers.
– André Nicolas
Aug 2 '15 at 18:56
By the way, if $a_k$ is the number of such sequences of length $k$, one can show quite easily (no binomial coefficients) that $a_n=a_n-1+a_n-2$. It is easy to find $a_1$ and $a_2$. Now use the recurrence to work your way up to $10$. We get the familiar Fibonacci numbers.
– André Nicolas
Aug 2 '15 at 18:56
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your formula is right. Your algebra is wrong.
$$1+ binom64 + binom73 +binom82 +binom91 +1=89$$
Try redoing the calculation and seeing. Link.
answered Aug 2 '15 at 18:42
Jahan Claes
95738
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
Your formula is right. Your algebra is wrong.
$$1+ binom64 + binom73 +binom82 +binom91 +1=89$$
Try redoing the calculation and seeing. Link.
answered Aug 2 '15 at 18:42
Jahan Claes
95738
add a comment |Â
up vote
1
down vote
accepted
Your formula is right. Your algebra is wrong.
$$1+ binom64 + binom73 +binom82 +binom91 +1=89$$
Try redoing the calculation and seeing. Link.
answered Aug 2 '15 at 18:42
Jahan Claes
95738
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your formula is right. Your algebra is wrong.
$$1+ binom64 + binom73 +binom82 +binom91 +1=89$$
Try redoing the calculation and seeing. Link.
answered Aug 2 '15 at 18:42
Jahan Claes
95738
Your formula is right. Your algebra is wrong.
$$1+ binom64 + binom73 +binom82 +binom91 +1=89$$
Try redoing the calculation and seeing. Link.
answered Aug 2 '15 at 18:42
Jahan Claes
95738
answered Aug 2 '15 at 18:42
Jahan Claes
95738
answered Aug 2 '15 at 18:42
Jahan Claes
95738
answered Aug 2 '15 at 18:42
Jahan Claes
95738
95738
add a comment |Â
add a comment |Â
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Check your sum again. I got $89$ using the same method you used.
– lulu
Aug 2 '15 at 18:42
Oops! That was so silly,posting it here. Answer did not match and I thought formula was wrong.
– user118494
Aug 2 '15 at 18:45
No worries. Combinatorics are simply confusing.
– lulu
Aug 2 '15 at 18:49
2
By the way, if $a_k$ is the number of such sequences of length $k$, one can show quite easily (no binomial coefficients) that $a_n=a_n-1+a_n-2$. It is easy to find $a_1$ and $a_2$. Now use the recurrence to work your way up to $10$. We get the familiar Fibonacci numbers.
– André Nicolas
Aug 2 '15 at 18:56