# Math Help - Multinomial question....

1. ## Multinomial question....

Hi there just going through some old exam papers preparing for the summer and i have come across this multinomial question. We have gone through this at university but very quickly, i understand the method, but this question is a bit harder than what i have done, i have had a go at answering it so this is what i have:
Question
F(X,Y,Z)=(X+2Y^2+Z)^4 AND G(X,Y,Z)=(X-Y+5Z)^6
Determine the coefficients with which the following terms appear in F and G

1) X^2*Z^2
For F I let: a=x b=2y^2 and c=z
Therefore forming:
F(a,b,c)=(a+b+c)^4
Hence x^2*z^2=a^2*c^2
=4!/(2!2!)=6

1) Again X^2*Z^2 had to be worked out for G
For G I let: a=x b=-y c=5z
G(a,b,c)=(a+b+c)^6
X^2*Z^2=a^2*c^2
=(6!/(2!2!))*25=4500?

2) x*y^4*z
For F
x*y^4*z=a*b^2*C
=(4!/(1!1!2!))*4=48?

2) x*y^4*z
For g
x*y^4*z=a*b^4*c
=(6!/(1!1!4!)*(1*5)=150

3) x*y^3*z^2
For f i havent got a clue im not sure if its possible

3) x*y^4*z
For G
x*y^4*z=a*b^3*z^2
=(6!/(3!*2!)*(-1*25)=-1500

Many thanks in advance for any help that can be given.

2. The polinomial formula is...

$\displaystyle (x_{1} + x_{2} + ... + x_{p})^{n} = \sum \frac{n!}{n_{1}!\ n_{2}! ... n_{p}! }\ x_{1}^{n_{1}}\ x_{2}^{n_{2}}\ ... x_{p}^{n_{p}}$ (1)

.... where $\Sigma$ is extended to all non negative integers $n_{1}, n_{2}, ..., n_{p}$ such that is $n_{1} + n_{2} +...+ n_{p}= n$ . Although Leibnitz 'discovered' this formula in seventeenth century, even today not all are really 'enthusiastic' of it...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
The polinomial formula is...

$\displaystyle (x_{1} + x_{2} + ... + x_{p})^{n} = \sum \frac{n!}{n_{1}!\ n_{2}! ... n_{p}! }\ x_{1}^{n_{1}}\ x_{2}^{n_{2}}\ ... x_{p}^{n_{p}}$ (1)

.... where $\Sigma$ is extended to all non negative integers $n_{1}, n_{2}, ..., n_{p}$ such that is $n_{1} + n_{2} +...+ n_{p}= n$ . Although Leibnitz 'discovered' this formula in seventeenth century, even today not all are really 'enthusiastic' of it...

Kind regards

$\chi$ $\sigma$
Many thanks for the reply we were given this in one of our lectures, just a matter of putting it in practice, we were only given simple examples such as:
f(x,y,z)=(x+y^2+z) find x*y^4*z etc fairly straight forward where a simple change can be used, hence none of the example contained a constant other than 1 in the multinomial, hence the questions in the paper are examples of what im looking at.

4. Anyone out there that can help me? many thanks.

5. Still looking for a bit of help, would be much appreciated.

6. Originally Posted by breitling
Still looking for a bit of help, would be much appreciated.
1f) 4!/(2!2!) is right
1g) it is 0 because 2+2 does not equal 6

2f) Note that y^4 is (y^2)^2 so this coefficient is nonzero since 1+2+1 = 4. You have it right, but it helps to write 2^2 instead of just 4, for clarity, as in 4!/(1!2!1!) * 2^2 = 48.
2g) 150 is right, again you can write more clearly as 6!/(1!4!1!) * (-1)^4 * 5^1 = 150.

I've helped you with the first two... you should be more confident about whether your third solution is right or wrong now..

Note that you can use CAS's to easily check your answers; Pari/GP is freeware (open source, in fact) and very good, Wolfram Alpha is online and also free.

7. Originally Posted by undefined
1f) 4!/(2!2!) is right
1g) it is 0 because 2+2 does not equal 6

2f) Note that y^4 is (y^2)^2 so this coefficient is nonzero since 1+2+1 = 4. You have it right, but it helps to write 2^2 instead of just 4, for clarity, as in 4!/(1!2!1!) * 2^2 = 48.
2g) 150 is right, again you can write more clearly as 6!/(1!4!1!) * (-1)^4 * 5^1 = 150.

I've helped you with the first two... you should be more confident about whether your third solution is right or wrong now..

Note that you can use CAS's to easily check your answers; Pari/GP is freeware (open source, in fact) and very good, Wolfram Alpha is online and also free.
Many thanks think i have the hang of it now thanks once more.