Expand the trinomial $(x+y+z)^4$ using the Multinomial Theorem
$begingroup$
Use the multinomial theorem to expand $(x+y+z)^4$.
To calculate the number of terms, you apply the following formula: $binom{n+r-1}{n}$. Here $n=4$ and $r=3$. So $binom{6}{4}=15$. I don't understand how they are getting $15$ terms.
The Multinomial Theorem states:
$(x_1+cdots+x_r)^n = sumlimits_{n_1+...+n_r = n}binom{n}{n_1,...,n_r}cdot x_1^{n_1} cdots x_r^{n_r}$.
The sum runs over all possible partitions $n_1,...,n_r$ such that $n_1+...+n_r=n$
There are only $3$ sets of numbers to get $n=4$:
1. $(4,0,0) implies$ Here there are $3$ objects but $2$ are the same. So there are $3!/2! = 3$ permutations
2. $(3,1,0) implies 3! = 6$ Permutations
3. $(2,2,0) implies 3!/2! = 3$
$3+6+3 = 12$.
So for example, for item one you would have:
$binom{4}{4,0,0} x^4 y^0 z^0 + binom{4}{0,4,0} x^0 y^4 z^0 + binom{4}{0,0,4} x^0 y^0 z^4$
For item $2\$:
$(3,0,1), (3,1,0), \
(1,3,0), (0,3,1), \
(1,0,3), (0,1,3)$
For item $3\$:
$(2,2,0), (0,2,2,), (2,0,2)$
What are the other $3$ terms? Thank you!
multinomial-coefficients
edited Jan 3 at 9:52
E.Nole
166114
asked May 7 '13 at 19:59
user1527227user1527227
94822548
$endgroup$
add a comment |
$begingroup$
Use the multinomial theorem to expand $(x+y+z)^4$.
To calculate the number of terms, you apply the following formula: $binom{n+r-1}{n}$. Here $n=4$ and $r=3$. So $binom{6}{4}=15$. I don't understand how they are getting $15$ terms.
The Multinomial Theorem states:
$(x_1+cdots+x_r)^n = sumlimits_{n_1+...+n_r = n}binom{n}{n_1,...,n_r}cdot x_1^{n_1} cdots x_r^{n_r}$.
The sum runs over all possible partitions $n_1,...,n_r$ such that $n_1+...+n_r=n$
There are only $3$ sets of numbers to get $n=4$:
1. $(4,0,0) implies$ Here there are $3$ objects but $2$ are the same. So there are $3!/2! = 3$ permutations
2. $(3,1,0) implies 3! = 6$ Permutations
3. $(2,2,0) implies 3!/2! = 3$
$3+6+3 = 12$.
So for example, for item one you would have:
$binom{4}{4,0,0} x^4 y^0 z^0 + binom{4}{0,4,0} x^0 y^4 z^0 + binom{4}{0,0,4} x^0 y^0 z^4$
For item $2\$:
$(3,0,1), (3,1,0), \
(1,3,0), (0,3,1), \
(1,0,3), (0,1,3)$
For item $3\$:
$(2,2,0), (0,2,2,), (2,0,2)$
What are the other $3$ terms? Thank you!
multinomial-coefficients
edited Jan 3 at 9:52
E.Nole
166114
asked May 7 '13 at 19:59
user1527227user1527227
94822548
$endgroup$
2
$begingroup$
I think you forgot about $(1,1,2)$?
$endgroup$
– Suugaku
May 7 '13 at 20:03
$begingroup$
Yeah. that's it. Thanks!
$endgroup$
– user1527227
May 7 '13 at 20:05
add a comment |
$begingroup$
Use the multinomial theorem to expand $(x+y+z)^4$.
To calculate the number of terms, you apply the following formula: $binom{n+r-1}{n}$. Here $n=4$ and $r=3$. So $binom{6}{4}=15$. I don't understand how they are getting $15$ terms.
The Multinomial Theorem states:
$(x_1+cdots+x_r)^n = sumlimits_{n_1+...+n_r = n}binom{n}{n_1,...,n_r}cdot x_1^{n_1} cdots x_r^{n_r}$.
The sum runs over all possible partitions $n_1,...,n_r$ such that $n_1+...+n_r=n$
There are only $3$ sets of numbers to get $n=4$:
1. $(4,0,0) implies$ Here there are $3$ objects but $2$ are the same. So there are $3!/2! = 3$ permutations
2. $(3,1,0) implies 3! = 6$ Permutations
3. $(2,2,0) implies 3!/2! = 3$
$3+6+3 = 12$.
So for example, for item one you would have:
$binom{4}{4,0,0} x^4 y^0 z^0 + binom{4}{0,4,0} x^0 y^4 z^0 + binom{4}{0,0,4} x^0 y^0 z^4$
For item $2\$:
$(3,0,1), (3,1,0), \
(1,3,0), (0,3,1), \
(1,0,3), (0,1,3)$
For item $3\$:
$(2,2,0), (0,2,2,), (2,0,2)$
What are the other $3$ terms? Thank you!
multinomial-coefficients
edited Jan 3 at 9:52
E.Nole
166114
asked May 7 '13 at 19:59
user1527227user1527227
94822548
$endgroup$
Use the multinomial theorem to expand $(x+y+z)^4$.
To calculate the number of terms, you apply the following formula: $binom{n+r-1}{n}$. Here $n=4$ and $r=3$. So $binom{6}{4}=15$. I don't understand how they are getting $15$ terms.
The Multinomial Theorem states:
$(x_1+cdots+x_r)^n = sumlimits_{n_1+...+n_r = n}binom{n}{n_1,...,n_r}cdot x_1^{n_1} cdots x_r^{n_r}$.
The sum runs over all possible partitions $n_1,...,n_r$ such that $n_1+...+n_r=n$
There are only $3$ sets of numbers to get $n=4$:
1. $(4,0,0) implies$ Here there are $3$ objects but $2$ are the same. So there are $3!/2! = 3$ permutations
2. $(3,1,0) implies 3! = 6$ Permutations
3. $(2,2,0) implies 3!/2! = 3$
$3+6+3 = 12$.
So for example, for item one you would have:
$binom{4}{4,0,0} x^4 y^0 z^0 + binom{4}{0,4,0} x^0 y^4 z^0 + binom{4}{0,0,4} x^0 y^0 z^4$
For item $2\$:
$(3,0,1), (3,1,0), \
(1,3,0), (0,3,1), \
(1,0,3), (0,1,3)$
For item $3\$:
$(2,2,0), (0,2,2,), (2,0,2)$
What are the other $3$ terms? Thank you!
multinomial-coefficients
multinomial-coefficients
edited Jan 3 at 9:52
E.Nole
166114
asked May 7 '13 at 19:59
user1527227user1527227
94822548
edited Jan 3 at 9:52
E.Nole
166114
asked May 7 '13 at 19:59
user1527227user1527227
94822548
edited Jan 3 at 9:52
E.Nole
166114
edited Jan 3 at 9:52
E.Nole
166114
edited Jan 3 at 9:52
E.Nole
166114
166114
asked May 7 '13 at 19:59
user1527227user1527227
94822548
asked May 7 '13 at 19:59
user1527227user1527227
94822548
asked May 7 '13 at 19:59
user1527227user1527227
94822548
94822548
2
$begingroup$
I think you forgot about $(1,1,2)$?
$endgroup$
– Suugaku
May 7 '13 at 20:03
$begingroup$
Yeah. that's it. Thanks!
$endgroup$
– user1527227
May 7 '13 at 20:05
add a comment |
2
$begingroup$
I think you forgot about $(1,1,2)$?
$endgroup$
– Suugaku
May 7 '13 at 20:03
$begingroup$
Yeah. that's it. Thanks!
$endgroup$
– user1527227
May 7 '13 at 20:05
2
2
$begingroup$
I think you forgot about $(1,1,2)$?
$endgroup$
– Suugaku
May 7 '13 at 20:03
$begingroup$
I think you forgot about $(1,1,2)$?
$endgroup$
– Suugaku
May 7 '13 at 20:03
$begingroup$
Yeah. that's it. Thanks!
$endgroup$
– user1527227
May 7 '13 at 20:05
$begingroup$
Yeah. that's it. Thanks!
$endgroup$
– user1527227
May 7 '13 at 20:05
add a comment |
1 Answer
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$begingroup$
Perhaps the other three terms that you are missing correspond to (1,1,2)?
answered May 7 '13 at 20:03
user76539user76539
1162
$endgroup$
add a comment |
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$begingroup$
Perhaps the other three terms that you are missing correspond to (1,1,2)?
answered May 7 '13 at 20:03
user76539user76539
1162
$endgroup$
add a comment |
$begingroup$
Perhaps the other three terms that you are missing correspond to (1,1,2)?
answered May 7 '13 at 20:03
user76539user76539
1162
$endgroup$
add a comment |
$begingroup$
Perhaps the other three terms that you are missing correspond to (1,1,2)?
answered May 7 '13 at 20:03
user76539user76539
1162
$endgroup$
Perhaps the other three terms that you are missing correspond to (1,1,2)?
answered May 7 '13 at 20:03
user76539user76539
1162
answered May 7 '13 at 20:03
user76539user76539
1162
answered May 7 '13 at 20:03
user76539user76539
1162
answered May 7 '13 at 20:03
user76539user76539
1162
1162
add a comment |
add a comment |
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2
$begingroup$
I think you forgot about $(1,1,2)$?
$endgroup$
– Suugaku
May 7 '13 at 20:03
$begingroup$
Yeah. that's it. Thanks!
$endgroup$
– user1527227
May 7 '13 at 20:05