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[[Category:MA375Spring2009Walther]] | [[Category:MA375Spring2009Walther]] | ||
| + | '''Back to [[HW 4: 5.5]]''' | ||
Can someone explain to me how i would go about this problem? | Can someone explain to me how i would go about this problem? | ||
| − | Here's how I thought about it. After all terms with identical sets of exponents are added, each term looks like this: <math> x_{1}^{?}*x_{2}^{?}*...*x_{m}^{?} </math>. All the exponents have to add up to n, and there are m x's on which to place exponents. So, the question is simply a case of partitioning n identical items into m distinguishable groups. How many different terms are there? I think the answer is | + | Here's how I thought about it. After all terms with identical sets of exponents are added, each term looks like this: <math> x_{1}^{?}*x_{2}^{?}*...*x_{m}^{?} </math>. All the exponents have to add up to n, and there are m x's on which to place exponents. So, the question is simply a case of partitioning n identical items into m distinguishable groups. How many different terms are there? I think the answer is <math>{n+m-1 \choose m-1}</math>, but I don't know how to write it in Latex. =P Anyway, I hope this helps. -Zoe |
I found that it was a lot easier to think of it with real numbers..like (x+y+z)^7...so, m=3 and n=7. then figure it out with the numbers and plug it into a general equation. -Brandy | I found that it was a lot easier to think of it with real numbers..like (x+y+z)^7...so, m=3 and n=7. then figure it out with the numbers and plug it into a general equation. -Brandy | ||
Latest revision as of 20:59, 11 February 2009
Back to HW 4: 5.5
Can someone explain to me how i would go about this problem?
Here's how I thought about it. After all terms with identical sets of exponents are added, each term looks like this: $ x_{1}^{?}*x_{2}^{?}*...*x_{m}^{?} $. All the exponents have to add up to n, and there are m x's on which to place exponents. So, the question is simply a case of partitioning n identical items into m distinguishable groups. How many different terms are there? I think the answer is $ {n+m-1 \choose m-1} $, but I don't know how to write it in Latex. =P Anyway, I hope this helps. -Zoe
I found that it was a lot easier to think of it with real numbers..like (x+y+z)^7...so, m=3 and n=7. then figure it out with the numbers and plug it into a general equation. -Brandy
