If i have 90 people coming to an event, with 120 different outfits, what is the chance of two people wearing matching outfits
How can 90 people wear 120 different outfits?
I guess you mean that 90 people come and they randomly choose 1 of 120 outfits each.
Let me see...
The number of ways that people can have different outfits (order counts) is
$$120P90 = 120!/30! \approx 2.5219*10^{166}$$
The number of ways that they can have any outfit = $$120^{90}=1.3376\times 10^{187}$$
The prob that all the outfits are different would be
$$\frac{120!}{30!} \div 120^{90} \approx 1.8855\times 10^{-21}$$
the prob that 2 or more are in the same outfit would be
$$\approx 1-{ 1.8855\times 10^{-21}}=1$$ correct to more than 10 decimal places.
I am not sure of this answer - repeat at own risk. LOL
How can 90 people wear 120 different outfits?
I guess you mean that 90 people come and they randomly choose 1 of 120 outfits each.
Let me see...
The number of ways that people can have different outfits (order counts) is
$$120P90 = 120!/30! \approx 2.5219*10^{166}$$
The number of ways that they can have any outfit = $$120^{90}=1.3376\times 10^{187}$$
The prob that all the outfits are different would be
$$\frac{120!}{30!} \div 120^{90} \approx 1.8855\times 10^{-21}$$
the prob that 2 or more are in the same outfit would be
$$\approx 1-{ 1.8855\times 10^{-21}}=1$$ correct to more than 10 decimal places.
I am not sure of this answer - repeat at own risk. LOL