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In a 5 digit ID number, what is the probability of exactly t

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Re: ID number. [#permalink] New post 04 Mar 2010, 22:24
The answer is simple
Step 1: Total number of ways = 10*10*10*10*10 = 10^5
Step 2: First choose three places for three 2s which can be done in 5C3 ways. (No point arranging these 2s amongst themselves) = 10 ways
Step 3: Now for the remaining two positions there are 9 numbers (as 2 is already used) each to be placed which can be done in 9*9 ways.
Step 4: favorable ways = 10*9*9 = 810

Thus probability = favorable ways / total ways = 810/10^5=0.0081 or 0.81%

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Re: ID number. [#permalink] New post 01 Jun 2014, 06:30
X X X X X

5C3 = 10, 10 3 place combination which can take 2.

if first 3 places have taken 2 as 2 2 2 _ _ then remaining places can take values between 0 -9 except 2. i.e 9.

therefore all possible such IDs 10*9*9 = 810

Overall possible combinations for 5 digit ID = 10 ^5

Therefore \frac{810}{10^5} * 100 = 0.81%. Ans A
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Re: ID number.   [#permalink] 01 Jun 2014, 06:30
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