Mixing
Binomial Theorem with Three Terms
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$(x^2 + 2 + frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
combinatorics binomial-coefficients
asked 17 hours ago
ultralight
325
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up vote
4
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favorite
$(x^2 + 2 + frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
combinatorics binomial-coefficients
asked 17 hours ago
ultralight
325
Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
– Henry
15 hours ago
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up vote
4
down vote
favorite
up vote
4
down vote
favorite
$(x^2 + 2 + frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
combinatorics binomial-coefficients
asked 17 hours ago
ultralight
325
$(x^2 + 2 + frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
combinatorics binomial-coefficients
combinatorics binomial-coefficients
asked 17 hours ago
ultralight
325
asked 17 hours ago
ultralight
325
asked 17 hours ago
ultralight
325
asked 17 hours ago
ultralight
325
asked 17 hours ago
ultralight
325
325
Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
– Henry
15 hours ago
add a comment |
Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
– Henry
15 hours ago
Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
– Henry
15 hours ago
Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
– Henry
15 hours ago
add a comment |
5 Answers
5
active
oldest
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up vote
11
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In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$
There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.
Thus the coefficient of $x^8$ is $8(35)+21 = 301$
answered 17 hours ago
Mohammad Riazi-Kermani
40.1k41958
But in the first type, don't you also have to multiply by $(x^{-1})^3$?
– mathnoob
17 hours ago
@mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
– Mohammad Riazi-Kermani
17 hours ago
add a comment |
up vote
8
down vote
The multinomial theorem can come to the rescue:
$$
(a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
$$
where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.
Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
$$
2i-k=8,qquad i+kle 7
$$
Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases
$i=4$, $k=0$, $j=3$;
$i=5$, $k=2$, $j=0$.
Thus the coefficient is
$$
2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
$$
answered 12 hours ago
egreg
173k1383197
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up vote
5
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Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.
Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.
If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$
If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$
If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$
so the answer is $301$.
edited 12 hours ago
Rad80
30718
answered 17 hours ago
greedoid
34.3k114488
add a comment |
up vote
2
down vote
The answer is 301.
Just trust your plan of the twofold use of the binomial formula:
First step
$$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$
Second step
$$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$
Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$
$$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$
Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.
answered 17 hours ago
Dr. Wolfgang Hintze
3,020516
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0
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To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.
This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have
7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7
7+4+4, 4+7+4, 4+4+7
6+6+3, 6+3+6, 3+6+6
6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6
That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).
answered 9 hours ago
Acccumulation
6,4402616
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$
There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.
Thus the coefficient of $x^8$ is $8(35)+21 = 301$
answered 17 hours ago
Mohammad Riazi-Kermani
40.1k41958
But in the first type, don't you also have to multiply by $(x^{-1})^3$?
– mathnoob
17 hours ago
@mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
– Mohammad Riazi-Kermani
17 hours ago
add a comment |
up vote
11
down vote
accepted
In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$
There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.
Thus the coefficient of $x^8$ is $8(35)+21 = 301$
answered 17 hours ago
Mohammad Riazi-Kermani
40.1k41958
But in the first type, don't you also have to multiply by $(x^{-1})^3$?
– mathnoob
17 hours ago
@mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
– Mohammad Riazi-Kermani
17 hours ago
add a comment |
up vote
11
down vote
accepted
up vote
11
down vote
accepted
In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$
There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.
Thus the coefficient of $x^8$ is $8(35)+21 = 301$
answered 17 hours ago
Mohammad Riazi-Kermani
40.1k41958
In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$
There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.
Thus the coefficient of $x^8$ is $8(35)+21 = 301$
answered 17 hours ago
Mohammad Riazi-Kermani
40.1k41958
answered 17 hours ago
Mohammad Riazi-Kermani
40.1k41958
answered 17 hours ago
Mohammad Riazi-Kermani
40.1k41958
answered 17 hours ago
Mohammad Riazi-Kermani
40.1k41958
40.1k41958
But in the first type, don't you also have to multiply by $(x^{-1})^3$?
– mathnoob
17 hours ago
@mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
– Mohammad Riazi-Kermani
17 hours ago
add a comment |
But in the first type, don't you also have to multiply by $(x^{-1})^3$?
– mathnoob
17 hours ago
@mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
– Mohammad Riazi-Kermani
17 hours ago
But in the first type, don't you also have to multiply by $(x^{-1})^3$?
– mathnoob
17 hours ago
But in the first type, don't you also have to multiply by $(x^{-1})^3$?
– mathnoob
17 hours ago
@mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
– Mohammad Riazi-Kermani
17 hours ago
@mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
– Mohammad Riazi-Kermani
17 hours ago
add a comment |
up vote
8
down vote
The multinomial theorem can come to the rescue:
$$
(a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
$$
where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.
Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
$$
2i-k=8,qquad i+kle 7
$$
Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases
$i=4$, $k=0$, $j=3$;
$i=5$, $k=2$, $j=0$.
Thus the coefficient is
$$
2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
$$
answered 12 hours ago
egreg
173k1383197
add a comment |
up vote
8
down vote
The multinomial theorem can come to the rescue:
$$
(a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
$$
where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.
Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
$$
2i-k=8,qquad i+kle 7
$$
Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases
$i=4$, $k=0$, $j=3$;
$i=5$, $k=2$, $j=0$.
Thus the coefficient is
$$
2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
$$
answered 12 hours ago
egreg
173k1383197
add a comment |
up vote
8
down vote
up vote
8
down vote
The multinomial theorem can come to the rescue:
$$
(a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
$$
where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.
Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
$$
2i-k=8,qquad i+kle 7
$$
Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases
$i=4$, $k=0$, $j=3$;
$i=5$, $k=2$, $j=0$.
Thus the coefficient is
$$
2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
$$
answered 12 hours ago
egreg
173k1383197
The multinomial theorem can come to the rescue:
$$
(a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
$$
where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.
Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
$$
2i-k=8,qquad i+kle 7
$$
Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases
$i=4$, $k=0$, $j=3$;
$i=5$, $k=2$, $j=0$.
Thus the coefficient is
$$
2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
$$
answered 12 hours ago
egreg
173k1383197
edited 7 hours ago
answered 12 hours ago
egreg
173k1383197
answered 12 hours ago
egreg
173k1383197
answered 12 hours ago
egreg
173k1383197
173k1383197
add a comment |
add a comment |
up vote
5
down vote
Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.
Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.
If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$
If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$
If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$
so the answer is $301$.
edited 12 hours ago
Rad80
30718
answered 17 hours ago
greedoid
34.3k114488
add a comment |
up vote
5
down vote
Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.
Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.
If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$
If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$
If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$
so the answer is $301$.
edited 12 hours ago
Rad80
30718
answered 17 hours ago
greedoid
34.3k114488
add a comment |
up vote
5
down vote
up vote
5
down vote
Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.
Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.
If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$
If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$
If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$
so the answer is $301$.
edited 12 hours ago
Rad80
30718
answered 17 hours ago
greedoid
34.3k114488
Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.
Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.
If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$
If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$
If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$
so the answer is $301$.
edited 12 hours ago
Rad80
30718
answered 17 hours ago
greedoid
34.3k114488
edited 12 hours ago
Rad80
30718
edited 12 hours ago
Rad80
30718
edited 12 hours ago
Rad80
30718
30718
answered 17 hours ago
greedoid
34.3k114488
answered 17 hours ago
greedoid
34.3k114488
answered 17 hours ago
greedoid
34.3k114488
34.3k114488
add a comment |
add a comment |
up vote
2
down vote
The answer is 301.
Just trust your plan of the twofold use of the binomial formula:
First step
$$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$
Second step
$$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$
Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$
$$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$
Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.
answered 17 hours ago
Dr. Wolfgang Hintze
3,020516
add a comment |
up vote
2
down vote
The answer is 301.
Just trust your plan of the twofold use of the binomial formula:
First step
$$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$
Second step
$$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$
Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$
$$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$
Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.
answered 17 hours ago
Dr. Wolfgang Hintze
3,020516
add a comment |
up vote
2
down vote
up vote
2
down vote
The answer is 301.
Just trust your plan of the twofold use of the binomial formula:
First step
$$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$
Second step
$$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$
Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$
$$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$
Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.
answered 17 hours ago
Dr. Wolfgang Hintze
3,020516
The answer is 301.
Just trust your plan of the twofold use of the binomial formula:
First step
$$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$
Second step
$$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$
Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$
$$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$
Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.
answered 17 hours ago
Dr. Wolfgang Hintze
3,020516
edited 16 hours ago
answered 17 hours ago
Dr. Wolfgang Hintze
3,020516
answered 17 hours ago
Dr. Wolfgang Hintze
3,020516
answered 17 hours ago
Dr. Wolfgang Hintze
3,020516
3,020516
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To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.
This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have
7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7
7+4+4, 4+7+4, 4+4+7
6+6+3, 6+3+6, 3+6+6
6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6
That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).
answered 9 hours ago
Acccumulation
6,4402616
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To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.
This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have
7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7
7+4+4, 4+7+4, 4+4+7
6+6+3, 6+3+6, 3+6+6
6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6
That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).
answered 9 hours ago
Acccumulation
6,4402616
add a comment |
up vote
0
down vote
up vote
0
down vote
To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.
This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have
7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7
7+4+4, 4+7+4, 4+4+7
6+6+3, 6+3+6, 3+6+6
6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6
That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).
answered 9 hours ago
Acccumulation
6,4402616
To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.
This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have
7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7
7+4+4, 4+7+4, 4+4+7
6+6+3, 6+3+6, 3+6+6
6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6
That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).
answered 9 hours ago
Acccumulation
6,4402616
answered 9 hours ago
Acccumulation
6,4402616
answered 9 hours ago
Acccumulation
6,4402616
answered 9 hours ago
Acccumulation
6,4402616
6,4402616
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Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
– Henry
15 hours ago