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winnerhere: Master | Next Rank: 500 Posts; Posts: 231; Joined: Thu Apr 12, 2007 2:45 amFollowed by: 1 member; Thanked: 5 times

by winnerhere » Wed Aug 12, 2009 3:03 am

Friends,

May I know the answer for this sum.

"All the possible 5-digit numbers are formed using the digits 1,2,3,4 and 5 - repetition of digits being allowed.If one of these numbers is selected at random,what is the probability that it will have exactly one digit repeated and that too occuring twice"

1)343/625

2)48/125

3)49/125

4)none of these

I got the answer 12/125 and hence I think its none of these.

Can anyone help me with this sum

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Source: Beat The GMAT — Problem Solving | Wed Aug 12, 2009 3:03 am

ashis979: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Tue Jul 28, 2009 8:12 pm; Thanked: 4 times

by ashis979 » Wed Aug 12, 2009 6:10 am

I think the answer is 48/125.

Here's how I did it:

For the denominator,
Since repitition of digits is allowed, there are five ways to pick the first digit, five ways for the second digit,..., basically five ways to pick each of the five digits (1,2,3,4,5).
So, the total number of five-digit numbers you can have is: 5*5*5*5*5 = 3,125.

For the numerator,
Since the question is asking for the probability of picking a five-digit number that has only one digit repeated with two occurences, you can think of the number as, for example, a five digit number with two 1's:
11234 =>These digits can be arranged in 5! ways to form 5! = 120 numbers.
Similarly, it is also possible to have the following number with two 1's:
11235 (we didn't use the 5 above) => These digits as well, can be arranged in 5! = 120 ways (same as above).
So, in total you'll have 240 different five-digit numbers that have two 1's. Therefore, you'll have 240*5 (numbers with two 1's, or two 2's,..., or two 5's), for a total of 1,200 different five-digit numbers that have only one digit repeated two times.

So, the final probability is 1,200/3,125 = 48/125 (Option B).

-Ash

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real2008: Legendary Member; Posts: 527; Joined: Thu May 01, 2008 12:06 am; Thanked: 7 times

by real2008 » Wed Aug 12, 2009 6:16 am

I too find the answer to be 48/125.

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winnerhere: Master | Next Rank: 500 Posts; Posts: 231; Joined: Thu Apr 12, 2007 2:45 amFollowed by: 1 member; Thanked: 5 times

ashish

by winnerhere » Wed Aug 12, 2009 6:26 am

11234 =>These digits can be arranged in 5! ways to form 5! = 120 numbers.

no ashish.These digits cannot be arranged in 5! ways.They can be arranged only in 5!/2! ways

12341 is same as 12341!

Rule : The number of ways n digits can be arranged with r similar digits is n!/r!

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ben wade: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Sun Jun 14, 2009 4:54 am; Thanked: 1 times

by ben wade » Wed Aug 12, 2009 7:33 am

I go with None of these. The answer i got is 24/ 625.

Here is the logic i used -
Total number of ways - 5^5 = 3125

As for the number of ways to arrange the 5 digits in the 5-digit number with only one digit repeated with two occurences -

11--- the rest of the numbers can be filled in 4*3*2=24 ways.
for 22--- same as above - 24 ways.

Similarly for all the 5 numbers it is 24*5= 120 ways.

So the probabilility is 120/3125 = 24/625.
So none of these.
Let me know if this logic is incorrect.

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winnerhere: Master | Next Rank: 500 Posts; Posts: 231; Joined: Thu Apr 12, 2007 2:45 amFollowed by: 1 member; Thanked: 5 times

by winnerhere » Wed Aug 12, 2009 8:34 am

Guys by my logic its 48/625

just now checked it.

my logic,

case 1:

11 x y z - there are 4 options(2,3,4,5) for this 3 numbers x,y,z

there are 4c3 combinations of (2,3,4,5) = 4 combinations

for (1,1,2,3,4) we get 5!/2! arrangements = 60

so 60 * 4(different combinations) = 240 for (1,1,x,y,z)

for all 240 * 5 = 1200

1200/5*5*5*5*5* = 48/625

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prindaroy: Master | Next Rank: 500 Posts; Posts: 113; Joined: Thu Jul 16, 2009 11:23 am; Thanked: 15 times; GMAT Score:730

by prindaroy » Wed Aug 12, 2009 10:51 am

total poss. = 3125

11_ _ _ = 5!/2! * 4c3 = 240 (4c3 comes from the fact that you need to choose three from 2,3,4,5)

22_ _ _ = 5!/2! * 4c3 = 240

33_ _ _ = 5!/2! * 4c3 = 240

44_ _ _ = 5!/2! * 4c3 = 240

55_ _ _ = 5!/2! * 4c3 = 240

240*5 = 1200

Therefore; 1200/3125 = 48/125

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