If i have 90 people coming to an event, with 120 different outfits, what is the chance of two people wearing matching outfits

Guest Mar 24, 2015

#1**+5 **

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

Melody
Mar 25, 2015

#1**+5 **

Best Answer

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

Melody
Mar 25, 2015