# Homework Help: Binomial formula

1. Mar 13, 2012

### tony873004

1. The problem statement, all variables and given/known data
Find the term with the specified power in the expansion of the given binomial power.
$$\left( {x^3 + y^2 } \right)^{42} ,\,\,\,\,\,y^{15}$$

2. Relevant equations
$${\rm{term}} = \frac{{n!}}{{r!\left( {n - r} \right)!}}x^{n - r} y^r$$

3. The attempt at a solution
$$\begin{array}{l} {\rm{term}} = \frac{{42!}}{{15!\left( {42 - 15} \right)!}}x^{3 \cdot \left( {42 - 15} \right)} y^{2 \cdot 15} \\ \\ {\rm{term}} = \frac{{42!}}{{15!\left( {27} \right)!}}x^{81} y^{30} \\ {\rm{term}} = {\rm{98672427616}}\,x^{81} y^{30} \\ \end{array}$$

The back of the book says no such term exists. Why? Is it because x has an exponent that is higher than n? x^3 doesn't have a higher exponent, and I thought that's all that mattered.

Also, is there a way of simplifying that factorial so I don't have to rely completely on the calculator to solve? Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 13, 2012

### scurty

I would think a $y^{15}$ term does not exist because every $y$ term is raised to an even power? I've never used the formula before to calculate the term with the given binomial power so I can't comment on that part if you did it correctly.

For this factorial there isn't much else you can do except rewrite $\displaystyle\frac{42!}{15! \cdot 27!}$ as $\displaystyle\frac{(42 \cdot 41 \cdot \cdot \cdot 28)27!}{15! \cdot 27!} = \frac{(42 \cdot 41 \cdot \cdot \cdot 28)}{15!}$ and then cancel out like terms to get rid of the 15!.

3. Mar 13, 2012

### tony873004

Thanks! The longer I stared at this, I started to realize that there's nothing I can multiply by 2 that will give me 15. So I imagine that "does not exist" is also the answer to$$\left( {x^3 + y^2 } \right)^{42} ,\,\,\,\,\,y^{15}$$
This is an even question, so no back of book answer.

4. Mar 13, 2012

### scurty

Well, no number in the inters to multiply 2 by to get 15! :P

I think you recopied the problem from the first post, but did it involve another instance where you can't multiply an integer to get the specified power?

5. Mar 13, 2012

### tony873004

oops, my copy and paste skills need improving!
$$\left( {x^3 + y^2 } \right)^{107} ,\,\,\,\,y^{77}$$

6. Mar 13, 2012

### scurty

Yeah, same type of case.

7. Mar 13, 2012

### tony873004

Going back to the first example, if, rather than use the formula, I decide to actually waste a few sheets of paper expanding this this thing, my first few terms will be
$$x^{3\left( {42} \right)} y^{2\left( 0 \right)} + x^{3\left( {41} \right)} y^{2\left( 1 \right)} + x^{3\left( {40} \right)} y^{2\left( 2 \right)} + x^{3\left( {39} \right)} y^{2\left( 3 \right)} + ...$$
which is pretty much all I need to tell me that y^15 will never happen.

Thanks for the late night help!

8. Mar 13, 2012

### scurty

Don't forget to use pascal's triangle to put the correct coefficients in front of the terms! But, yep, that's why there is no term!