# Binomial expansion + interesting question

• Feb 1st 2011, 01:03 PM
fatlucky
Binomial expansion + interesting question
Hey,

how would I use a 'combinatoral proof' to prove this:

and here's a question that has got me puzzled..
A football team consists of 20 offensive and 20 defensive players. The players are to be paired in groups of 2 for the purpose of determining roommates.
(a) If the pairing is done at random, what is the probability that there are no offensive-defensive pairs?
(b) What is the probability that there are 2i offensive-defensive roommate pairs (with i= 1,2,...,10)

• Feb 1st 2011, 02:54 PM
harish21
a)
total of 40 players are grouped into two..

players can be sorted into ordered groups in $\displaystyle \dbinom{40}{2,2,....,2}=\dfrac{40!}{(2!)^{20}}$ ways.

now find the number of ways players can be sorted into unorderd pairs ......(I)

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also, a division will give you no offensive-defensive pairs if the offensive (and therefore the defensive) players are paired among themselves. So there are $\displaystyle \dfrac{20!}{10! \cdot 2^{10}}$ divisions for offensive as well as defensive players...(II)

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Using (I) and (II), you can now find the probability that there are no offensive-defensive pairs!
• Feb 2nd 2011, 02:42 AM
mr fantastic
Quote:

Originally Posted by fatlucky
Hey,

how would I use a 'combinatoral proof' to prove this:

[snip]