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Question

source: https://www.manhattangmat.com/index.cfm Code Blue A random 10-letter code is to be formed using the letters A, B, C, D, E, F, G, H, I and I (only the â€œIâ€

giddi77 · Answer

Wouldn't it be 1/5? 

All possible combinations = 10!/2! (need to divite by 2 since there are 2Is)

Required cominations where 2Is are together = 9!

Prob = 9!/(10!/2!)= 2*9!/10! = 1/5

Any commnets?

kook44 · Answer

I don't understand this - even though there are 2 I's, they each appear in the code once. What's the difference if 1 of the I's were a J (when computing the total number of possible combinations)? I would have said 10!
 
I just can't align this in my head. I'm trying to think of all the possible combinations:
II*******
*II******
**II*****
and so on up to 
*******II
I'm guessing the logic here is to treat the two I's together as one letter? so a nine letter combination = 9! ?

I would greatly appreciate an explanation. Want to train my brain to start thinking like this.

ywilfred · Answer

Number of 10 letter codes = 10!/2! = 

Number of 10 letter codes with the two "I" next to each other = 9!

P = 9!/(10!/2!) = (9! * 2!)/10! = 0.2 = 1/5

chiragr · Answer

Total combination = 10!/2!
Total combination II are to gather = 9!

9!/10!/2! = 1/5

But not very convincing!

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