Combinatorics

[beth2011]

I have been working on a homework set for over a week now. It is using the binomial theorem and Pascal triangle to expand (2s-3t^2)^6. I think the answer is 2s^6-12s^5(18t^2)+30s^4(45t^2)^2-40s^3(60t^2)^3+30s^2(45t^2)^5+(3t^2)^6.
However when I went on to try the next problem I was completely lost! This problem reads
What is the coefficient of x^20 in the expansion of (1+3x-x^2)(1+2x)^23? I am not even sure where to start on this one. I think we need to use the binomial theorem here as well I am just not sure how.
The last one that I really need help on reads "use the multinomial theorem to find the coefficient of x^22 in the expansion of (1+2x+x^3)^15. I have no clue what the multinomial theorem is or how to use it.
Please help I have been out of class for a week because my daughter was in the hospital and now I am LOST! PLEASE HELP :?

 

[Deleted member 4993]

I have been working on a homework set for over a week now. It is using the binomial theorem and Pascal triangle to expand (2s-3t^2)^6. I think the answer is 2s^6-12s^5(18t^2)+30s^4(45t^2)^2-40s^3(60t^2)^3+30s^2(45t^2)^5+(3t^2)^6.
However when I went on to try the next problem I was completely lost! This problem reads
What is the coefficient of x^20 in the expansion of (1+3x-x^2)(1+2x)^23? I am not even sure where to start on this one. I think we need to use the binomial theorem here as well I am just not sure how.
The last one that I really need help on reads "use the multinomial theorem to find the coefficient of x^22 in the expansion of (1+2x+x^3)^15. I have no clue what the multinomial theorem is or how to use it.
Please help I have been out of class for a week because my daughter was in the hospital and now I am LOST! PLEASE HELP :?

First find coefficient of x[sup:19ai2gh7]18[/sup:19ai2gh7] and x[sup:19ai2gh7]19[/sup:19ai2gh7] and x[sup:19ai2gh7]20[/sup:19ai2gh7] of (1+2x)[sup:19ai2gh7]23[/sup:19ai2gh7]

Now think ...

how those will help you to find coefficient of x[sup:19ai2gh7]20[/sup:19ai2gh7] in (1+3x-x[sup:19ai2gh7]2[/sup:19ai2gh7])(1+2x)[sup:19ai2gh7]23[/sup:19ai2gh7]?

.

 

[pmagunia]

Hello Beth.

If I understand correctly, you want to expand in the first problem: \(\displaystyle (2s-3t^2)^6\)

I got

\(\displaystyle 729\,t^{12}-2916\,s\,t^{10}+4860\,s^2\,t^8-4320\,s^3\,t^6+2160\,s^4%20\,t^4-576\,s^5\,t^2+64\,s^6$$\)

 

[soroban]

Hello, beth2011!

\(\displaystyle \text{Use the binomial theorem and Pascal's triangle to expand: }\2s-3t^2)^6\)

\(\displaystyle \text{I think the answer is: }\:2s^6-12s^5(18t^2)+30s^4(45t^2)^2-40s^3(60t^2)^3+30s^2(45t^2)^5+(3t^2)^6.\)
Sorry, this is wrong . . . besides, you haven't simplified it yet.

pmagunia's answer is correct . . . (I don't know why his/her answer is in reverse order.)

\(\displaystyle (2s - 3t^2)^6 \;=\;(2s)^6 \;-\; {6\choose5}(2x)^5(3t^2) \;+\; {6\choose4}(2s)^4(3t^2)^2 \;-\; {6\choose3}(2s)^3(3t^2)^3 \;+\; {6\choose2}(2s)^2(3t^2)^4 \;-\; {6\choose1}(2s)(3t^2)^5 \;+\; (3t^2)^6\)

. . . . . . . \(\displaystyle =\;64s^6 \;-\; 6\cdot32s^5\cdot3t^2 \;+\; 15\cdot16s^4\cdot9t^4 \;-\; 20\cdot8s^3\cdot27t^6 \;+\; 15\cdot4s^2\cdot81t^8 \;-\; 6\cdot2s\cdot243t^{10} \;+\; 729t^{12}\)

. . . . . . . \(\displaystyle =\;64s^6 - 576s^5t^2 + 2160s^4t^4 - 4320s^3t^6 + 4860s^2t^8 - 2916st^{10} + 729t^{12}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

\(\displaystyle \text{The Multinomial Theorem is quite complicated.}\)
. . \(\displaystyle \text{Fortunately, these problems involve only trinomials.}\)

\(\displaystyle \text{The expansion of }\,(a\,+\,b\,+\,c)^n\,\text{ contains }all\text{ the terms of the form: }\:{n\choose p,q,r}a^pb^qc^r\)
. . \(\displaystyle \text{where }\,p+q+r\:=\:n\:\text{ and }\,p,q,r\,\ge\,0\)


\(\displaystyle \text{I'll apply this to the last problem (it's easier).}\)

\(\displaystyle \text{We want the coefficient of }x^{22}\text{ in the expansion of }(x^3+2x+1)^{15}\)

\(\displaystyle \text{The terms are of the form: }\:{15\choose p,q,r}(x^3)^p(2x)^q(1)^r\:\text{ where }p+q+r \:=\:15\)

. . \(\displaystyle \text{And we want the terms which contain }x^{22}.\)


\(\displaystyle \text{This occurs in four cases:}\)

. . \(\displaystyle \begin{array}{cccccccc}{15\choose7,1,7}(x^3)^7(2x)^1(1)^7 &=& (51,\!470)(2)x^{22} &=& \;\;102,\!960x^{22} \\ \\[-3mm] {15\choose6,4,5}(x^3)^6(2x)^4(1)^5 &=& (180,\!180)(16)x^{22} &=& 2,\!882,\!880x^{22} \\ \\[-3mm] {15\choose5,7,3}(x^3)^5(2x)^7(1)^3 &=& (360,\!360)(128)x^{22} &=& 46,\!126,\!080x^{22} \\ \\[-3mm] {15\choose4,10,1}(x^3)^4(2x)^{10}(1)^1 &=& (15,\!015)(1024)x^{22} &=& 15,\!375,\!360x^{22} \\ \hline & & & & 64,\!486,\!280x^{22} \end{array}\)

\(\displaystyle \text{Therefore, the coefficient of }x^{22}\text{ is: }\:64,\!486,\!280.\)

But check my work . . . please!
.