Why 3?Follow that? Now consider the term resulting when j = 3.
Not sure I understand...How many ways can you get x^8 from (1+x^2-x^3)^9?
Note that every term in the expansion will be using exactly one term from each of the nine factors.
You can multiply 1*1*1*1*1*1x^2*(-x^3)(-x^3) = 1x^8. How many ways can you do that?
You can multiply 1*1*1*1*1*x^2*x^2*x^2*x^2 = 1x^8. How many ways can you do that?
Any other ways to get x^8?
I asked you to think. Whatever the expansion of [imath]\dbinom{9}{3} (-1)^3 * (1 + x^2)^6[/imath] may be, when you multiply it by [imath](x^3)^3 = x^9[/imath], how many terms involving [imath]x^8[/imath] are you going to get?Why 3?
I am afraid I don't get it.I asked you to think. Whatever the expansion of [imath]\dbinom{9}{3} (-1)^3 * (1 + x^2)^6[/imath] may be, when you multiply it by [imath](x^3)^3 = x^9[/imath], how many terms involving [imath]x^8[/imath] are you going to get?
It should quickly be obvious that if j = 3, every term in the expansion will be a multiple of a power of x of AT LEAST 9. If j = 4, every term in the expansion will be a multiple of a power of x of AT LEAST 12.
The first thing to do on some problem that is unfamiliar is to think rather than calculate.
How many ways can you get x^8 from (1+x^2-x^3)^9?
Note that every term in the expansion will be using exactly one term from each of the nine factors.
You can multiply 1*1*1*1*1*1x^2*(-x^3)(-x^3) = 1x^8. How many ways can you do that?
You can multiply 1*1*1*1*1*x^2*x^2*x^2*x^2 = 1x^8. How many ways can you do that?
Any other ways to get x^8?
When you multiply out (1+x^2-x^3)(1+x^2-x^3)(1+x^2-x^3)(1+x^2-x^3)(1+x^2-x^3)(1+x^2-x^3)(1+x^2-x^3)(1+x^2-x^3)(1+x^2-x^3) you pick one term from each factor to multiply.Not sure I understand...
Since Steve provided the answer, using my method:Using my results above, the answer would be 9!/(6!2!) + 9!/(5!4!).
I don't understand a lot of it unfortunately. First, where b1, b2, b3 come in? How do they relate to n and k.Since Steve provided the answer, using my method:
[math]{9\choose b_1,b_2,b_3}\underbrace{(1)^{b_1}(x^2)^{b_2}(-x^3)^{b_3}}_{\text{How to get }x^8?}[/math]Notice we achieve [imath]x^8[/imath], by letting [imath]b_1=5\,,b_2=4\,,b_3=0[/imath]
[math]{9\choose 5,4,0}(1)^{5}(x^2)^{4}(-x^3)^{0}=126x^8[/math]Or by letting [imath]b_1=6\,,b_2=1\,,b_3=2[/imath]
[math]{9\choose 6,1,2}(1)^{6}(x^2)^{1}(-x^3)^{2}=252x^8[/math]Therefore, the answer is
[math]\tag{which matches Steve's}{9\choose 5,4,0} +{9\choose 6,1,2} = \frac{9!}{5!4!0!} + \frac{9!}{6!2!1!}=378[/math]
Are you referring to the n and k in the binomial coefficient [imath]{n \choose k}?[/imath]I don't understand a lot of it unfortunately. First, where b1, b2, b3 come in? How do they relate to n and k.
YesAre you referring to the n and k in the binomial coefficient [imath]{n \choose k}?[/imath]
The binomial coefficient is a special case of the multinomial coefficient. As their names suggest, the prefix "bi-" deals with 2 groups, while the multinomial can deal with multiple groups, i.e. [imath]\ge 2[/imath].I don't understand a lot of it unfortunately. First, where b1, b2, b3 come in? How do they relate to n and k.
BBB, Please see post #2 where I stated the method which you used. Not a big deal. You can have the credit!Since Steve provided the answer, using my method:
[math]{9\choose b_1,b_2,b_3}\underbrace{(1)^{b_1}(x^2)^{b_2}(-x^3)^{b_3}}_{\text{How to get }x^8?}[/math]Notice we achieve [imath]x^8[/imath], by letting [imath]b_1=5\,,b_2=4\,,b_3=0[/imath]
[math]{9\choose 5,4,0}(1)^{5}(x^2)^{4}(-x^3)^{0}=126x^8[/math]Or by letting [imath]b_1=6\,,b_2=1\,,b_3=2[/imath]
[math]{9\choose 6,1,2}(1)^{6}(x^2)^{1}(-x^3)^{2}=252x^8[/math]Therefore, the answer is
[math]\tag{which matches Steve's}{9\choose 5,4,0} +{9\choose 6,1,2} = \frac{9!}{5!4!0!} + \frac{9!}{6!2!1!}=378[/math]
The early bird gets the worm, but the second mouse gets the cheeseBBB, Please see post #2 where I stated the method which you used. Not a big deal. You can have the credit!
Others have tried to explain the multinomial theorem to you. I avoid memorizing things that I seldom use. So here is an alternative way.I am afraid I don't get it.
Jeff,Others have tried to explain the multinomial theorem to you. I avoid memorizing things that I seldom use. So here is an alternative way.
[math]y = (1 + x^2 - x^3)^9 = (p - q)^9, \text { where } p = (x^0 + x^1) \text { and } q = x^3.[/math]
This is just a basic substitution of variables that will let us use the binomial theorem.
When we expand the binomial, we get
[math]y = (p - q)^9 = \sum_{j=0}^9 \dbinom{9}{j} (-1)^j p^{(9-j)} q^j.[/math]
Binomial theorem. No questions there I presume. But
[math]q = x^3 \implies q^3 = x^9, \ q^4 = x^{12}, \ q^5 = x^{15}, \text { etc.}[/math]
We are interested in the coefficient of [imath]x^8[/imath] in the expansion of y. But if j > 2, the terms will be multiples of powers of x of 9 or greater and so are irrelevant to powers of 8. The only terms that are relevant are when j = 0, j = 1, and j = 2. Let’s consider them one by one.
[math]j = 0 \implies \dbinom{9}{j} * (-1)^j p^{(9-j)}q^j = p^9 = (1 + x^2)^9 =\\ \sum_{k=0}^9 \dbinom{9}{k} 1^{(9-k)} (x^2)^k.[/math]
The only way we can get [imath]x^8[/imath] when j = 0 is when k = 4.
[math]\dbinom{9}{4} = \dfrac{9!}{4! * 5!} = \dfrac{9 * 8 * 7 * 6}{4 * 3 * 2} = 2 * 63 = 126.[/math]
Let’s think about j = 1.
[math]j = 1 \implies \dbinom{9}{j} (-1)^j p^{(9-j)}q^j = -9 p^8q = -9x^3(1 + x^2)^8.[/math]
Is it not obvious that multiplying an odd power of x by even powers of x will never generate [imath]x^8[/imath]?
j = 1 is irrelevant.
Let’s think about j = 2.
[math]j = 2 \implies \dbinom{9}{j} (-1)^j p^{(9-j)}q^j = 36x^6(1 + x^2)^7.[/math]
The only way to get [imath]x^8[/imath] out of that is with the term containing [imath]x^2[/imath].
[math](1 + x^2)^9 = \sum_{k=0}^7 \dbinom{7}{k}1^{(7-k)}(x^2)^k.[/math]
We need to look at k = 1. So the coefficient we want is 36 * 7 = 252.
Putting this altogether we get 252 + 126 = 378.
That is EXACTLY the number BBB got. His way is admittedly much more efficient. My way involves avoids memorizing another formula and is more easily explicable.