Coefficient in a polynomial
Task number: 4492
Variant
By the term \(xy^2\) and by \(xyz\) in the expansion of \((x+y+z)^3\).Solution
By multiplying-out the power \((x+y+z)^3=(x+y+z)(x+y+z)(x+y+z)\) we find that the term \(xy^2\) can be obtained in three ways: \(x\) is taken either from the first, the second or third factor of \((x+y+z)\) and the two terms \(y\) are taken from the other two.
Alternative way: each of the words \(xyy,yxy\) and \(yx\) corresponds to one of these possibilities. There are \(\frac{3!}{1!}{1!\cdot 2!}\) such words.
As above \(xyz\), we have three choices from which factor to choose \(x\) from, and then two choices of how to choose \(y\) from the other two. We then need to select \(z\) from the remaining term. That is \(3{\cdot} 2=6\) possibilities.
Here we build words from the three symbols \(x,y\) and \(z\), which can be done in \(3!\) ways.
Answer
The coefficient by \(xy^2\) is 3 and the coefficient by \(xyz\) is 6.Variant
By the term \(x^6 y^6 z^6\) in the expansion of \((x^2+3 y + 2 z^3)^{11}\).Solution
As in the previous variant, we first determine the number of 11-letter words in which the symbol \(x^2\) occurs three times, the symbol \(3y\) occurs six times, and the symbol \(2z^3\) occurs twice. One of the possible words, for example, is \(3y \cdot 3y \cdot 2z^3 \cdot x^2 \cdot x^2{} \cdot 3y \cdot 2z^3{} \cdot 3y \cdot 3y \cdot x^2 \). There are \(\frac{11!}{3!6!2!}=9240\) such words.
To get the coefficient by \(x^6 y^6 z^6\), we still need to extract the coefficient of \(3^6\) from the \((3y)^6\) and similarly \(2^2\) from the \((2z^3)^2\).
Answer
The coefficient by \(x^6 y^6 z^6\) is \(\frac{3^6 2^2 11!}{3!6!2!}=26 943 840\).
