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

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04 Mar 2010, 23:24
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%

Hope the solution helps
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01 Jun 2014, 07:30
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: In a 5 digit ID number, what is the probability of exactly t [#permalink]

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09 Aug 2015, 01:15
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Re: In a 5 digit ID number, what is the probability of exactly t [#permalink]

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05 May 2016, 06:08
Why have you done 5C3 and not 5!/3!. In my opinion the order matters so the 5!/3! should remove the three 2s arrangeemnt out of the equation

chetan2u wrote:
hi i think it does not req such long calculations my ans A...
there are 5 digits each place can have any of ten digits so total posb=10^5...
now three places only two can be there and rest two can have any of remaining nine digits...
so posb=1*1*1*9*9*5c3..
prob=.81%... although i might be still missing something.. ill try again
Re: In a 5 digit ID number, what is the probability of exactly t   [#permalink] 05 May 2016, 06:08

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