# Homework Help: Binomial Theorem

1. Mar 28, 2005

### Townsend

The questions are as follows

1) How many terms are there in the expansion of $$(a+3b-2c-d)^8$$ before like terms are combined?

$$4^8$$

2) How many terms are there after like terms are combined?

$$_8C_4=70$$

3) What is the coefficent of $$a^2b^3cd^2$$?

$$(_8C_2)( _6C_3)( _3C_1)( _2C_2) (3^3)(2^1)=90720$$

4) What is the coefficent of $$ac^4d^3$$?

$$(_8C_1)(_7C_4)(_3C_3)(2^4)=1225$$

5) In the expansion of $$(a+x+x^2)^5(a+x)^4$$, what is the coefficent of $$a^6x^3$$?

I don't really know where to start here but just to show everyone that I am trying this is what I have so far.

I know there are a total of 9 a's in all the factors. So I would have something like $$(_9C_6)(_3C_?)$$ where the ? is what I don't know for sure.

6) What is the coefficient of $$a^7x^5$$?

Same situation as above

I appericate any help. On the questions I have already answered I would like to know if your answers agree with mine or if I have made any mistakes. Thanks

Jeremy

2. Mar 28, 2005

### Townsend

For number 5 the answer is 84 I cheated and expanded the problem in maple. What I observed is that $$(_9C_3)(_6C_6)=84$$. Well the exponents add to 9 meaning that as far as the x's are concerned we only used the $$x^1's$$.

But number 6 is altogether different because the exponents add to 12 instead of 9 meaning that this one involves some of the $$x^2$$ terms. Well to start there are 7 a's so there are two other terms thus we must have two $$x^2$$ but that only works out to $$x^4$$ so I guess there cannot be a $$a^7x^5$$ term.

Well I guess with some thought and a little help from maple I found my own answers. But thanks for looking

Jeremy

Last edited: Mar 28, 2005
3. Mar 28, 2005

### xanthym

Answer #2 is not correct. Imagine 8 markers labeled "x" and 3 markers labeled "|". One possible combination would be:
1 Possible Arrangement ----> x|xxxx|xxx|
The "x" markers (if any) to the left of the leftmost "|" represent the power of "a", the "x" markers (if any) between the leftmost and middle "|" represent the power of "b", the "x" markers (if any) between the middle and rightmost "|" represent the power of "c", and finally the "x" markers (if any) to the right of the rightmost "|" represent the power of "d". (The above arrangement represents a1b4c3d0.)

Note that the sum of numbers of "x" markers is always 8, and that each of "a", "b', "c", & "d" can take appropriate power values {0 to 8}. Thus, the number of arrangements of 3 "|" among 8 "x" will equal the number of terms in the expansion when like terms are collected. That combinatorial number is:
Question #2:
{Number of Expansion Terms (with Like Terms Collected)} = (11!)/{(8!)(3!)} = (165)

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Last edited: Mar 29, 2005
4. Mar 29, 2005

### Townsend

I see....For some reason I have the hardest time seeing when to apply the whole slashes and x's method. What is the name of this method so that I can look for some problems online and get more practice with this method.

Thanks

5. Mar 29, 2005

### xanthym

This "x" & "|" technique illustrates the "Partition" method in combinatorics. Of course, it's characteristic of combinatorial problems that several different approaches can be applied to the same situation. In this case, the number of expansion terms (with like terms collected) can also be viewed like choosing combinations (order doesn't matter) of 8 from 4 with repetition, for which the formula is:

$$\ \ \ \ \color{red} \mathbb{C}_{\, (8)}^{\, (8 + 4 - 1)} \ = \ (11!)/{(8!)(3!)} = (165)$$

which is exactly the same previously derived.

~~

Last edited: Mar 30, 2005