N people sit down at random a classroom containing n+p seats

In summary, the question asks for the probability of all m red seats being occupied in a classroom with n+p total seats. One approach is to use the combination formula, with the numerator being the number of ways to choose m red seats and n-m non-red seats, and the denominator being the total number of ways to choose n people from n+p seats. Another approach is to first allocate the m red seats and then choose the remaining n-m people from the remaining n+p-m seats.
  • #1
indigojoker
246
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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 thinking n+p choose m since that will give the different ways that the red seats could be chosen, times n+p choose n-m which gives the choices that the non-red seats could be chosen.

Or: [tex] \frac{ C^{n+p} _{m} C^{n+p} _{n-m} } { C^{n+p} _{n} } [/tex]

Does this logic make sense?
 
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  • #2


indigojoker said:
I was thinking n+p choose m since that will give the different ways that the red seats could be chosen, times n+p choose n-m which gives the choices that the non-red seats could be chosen.
Close, but after having allocated m people amongst the n+p seats there are now only n+p-m seats to allocate the remaining n-m people.
 
  • #3


then:
[tex]
\frac{ C^{n+p} _{m} C^{n+p-m} _{n-m} } { C^{n+p} _{n} } [/tex]?
 
  • #4


indigojoker said:
then:
[tex]
\frac{ C^{n+p} _{m} C^{n+p-m} _{n-m} } { C^{n+p} _{n} } [/tex]?
EDIT: my error; see the other post.
 
Last edited:

1. What is the probability that all n people will have a seat in a classroom with n+p seats?

The probability can be calculated using the formula P(all n people have a seat) = (n+p)!/(p!*(n-1)!*n^p).

2. How does the number of seats in the classroom affect the probability of all n people having a seat?

The probability decreases as the number of seats (p) increases. This is because there are more possible combinations for people to sit in, making it less likely for all n people to have a seat.

3. Is it possible for more than n people to have a seat in a classroom with n+p seats?

Yes, it is possible for more than n people to have a seat. This can happen if some people choose to share seats or if the classroom has additional chairs or space for people to sit.

4. What is the maximum number of people that can have a seat in a classroom with n+p seats?

The maximum number of people that can have a seat is n+p, assuming that everyone chooses to sit down and there are no additional chairs or space in the classroom.

5. How does the arrangement of seats in the classroom affect the probability of all n people having a seat?

The arrangement of seats does not affect the probability. As long as there are enough seats for all n people to sit in, the probability remains the same. This is because the probability is based on the total number of possible combinations, not the specific arrangement of seats.

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