I'm still confused. If the above gives me the probability to seat those who are not in red chairs, then what happened to the probability to seat those who are IN red chairs? Does the question ask what is the probability to seat the red chairs INSTEAD of the probability to seat those who are NOT...
I thought I needed to find the probability of the red chairs being taken up?
If the red chairs are already taken up, then the numerator should be C^{n+p-m} _{n-m} }
But I'm not sure what this is getting at? The possible ways that the non-red chairs can be filled?
Ok, so the probability, I believe, to chose the red chairs should be:
\frac{ C^{n+p}_{m} }{C^{n+p}_{n} }
But I believe after we know how likely it is to choose the red seats, do we then multiply it by the number of people? Since there are n people choosing seats, then each would have an equal...
I don't think it's
\frac{ C^{n+p} _{m} C^{n+p-m} _{n-m} } { C^{n+p} _{n} }
The reason why is, I was told that in order for a relation like this to hold, you must have \frac{ C^{A} _{B} C^{D} _{F} } { C^{E} _{G} }
A+D=E
B+F=G
Also, I believe the denominator should be n+p choose m, since...
n people sit down at random a classroom containing n+p seats. There are m red seats (m<=n) in the classroom, what is the probability that all red seats will be occupied?
It is asking for a probability. The denominator is easy, I think it
should be (n+p) choose m since we are looking for the...
n people sit down at random a classroom containing n+p seats. There are m red seats (m<=n) in the classroom, what is the probability that all red seats will be occupied?
I know the bottom should be n+p choose n but I'm not sure what the numerator should be, any ideas would be great.
I was...
\sum^n _{j,k=0} (C^n _k x^{n-k})(C^n _j x^{j}) = \sum^{2n} _{k=0} C^{2n} _k x^{2n-k}
LHS: let j=k
RHS: let k=n
\sum^n _{k=0} (C^n _k C^n _k x^{n}) = \sum^{2n} _{n=0} C^{2n} _n x^{n}
for the polynomials to be equal, we must have the same coefficients, so can we say that:
since \sum_{n=0}...
Ah, I see the step that was skipped. Thanks for clarification.
\sum^n _{j,k=0} (C^n _k x^{n-k})(C^n _j x^{j}) = \sum^{2n} _{k=0} C^{2n} _k x^{2n-k}
We want the coefficient of each power to be the same since the polynomials are the same...
However, x^{n-k} x^{j} and x^{2n-k} do not...
Use Stirling's formula, n!=sqrt(2 pi n) n^n e^{-n}, to estimate the probability that all 50 states are represneted in a group of 50 councilmen chosen at random.
I think it should be:
P=\frac{50!}{50^{50}}
So using Stirling's formula, we get:
P=\frac{\sqrt{2 \pi 50} 50^{50} e^{-50}}{50^{50}}...
I'm still confused as to what was written.
I believe you meant to write:
\left( \sum^n _{k=0} C^n _k x^{n-k} \right) \left( \sum^n _{j=0} C^n _j x^{n-j} \right) = \sum^{2n} _{k=0} C^{2n} _k x^{2n-k}
I'm not sure what "The terms that produce a x^n" exactly means, which is why I don't...