Permutation Problem

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Permutation Problem

by akpareek » Mon Aug 09, 2010 12:02 pm
How many 3 digit number exist having exactly two 5's in them ?

A. 24 B. 26 C. 27 D. 29 E. 30

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by nicolezl » Mon Aug 09, 2010 12:15 pm
The possibilities are:

5 5 _

5 _ 5

_ 5 5

There are 9 possible values for each of the first two (all digits except 5), and 8 possible values for the last one (all digits except 5 and 0). So there are 26 3-digit numbers with exactly two 5s.

Pick B

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by clock60 » Mon Aug 09, 2010 12:16 pm
x55, 8 numbers ( x is any digit exept 0 and 5. 8*1*1)
5x5, 9 numbers ( x any number exept 5. 1*9*1)
55x, 9 numbers ( x any number exept 5 1*1*9)
total=8+9+9=26
my pick for B

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by indiantiger » Mon Aug 09, 2010 6:37 pm
total there will be three cases

55X -- in this case X can be replace with 9 other digits so 9 number of cases possible
5X5 -- in this case X can be replace with 9 other digits so 9 number of cases possible
X55 -- in this case X can be replace with 8 other digits so 8 number of cases possible {first digit cannot be 5 or 0}
9 + 9 + 8 = 26 (Answer)
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by zareentaj » Mon Aug 09, 2010 11:01 pm
indiantiger wrote:total there will be three cases

55X -- in this case X can be replace with 9 other digits so 9 number of cases possible
5X5 -- in this case X can be replace with 9 other digits so 9 number of cases possible
X55 -- in this case X can be replace with 8 other digits so 8 number of cases possible {first digit cannot be 5 or 0}
9 + 9 + 8 = 26 (Answer)
Thanks. indiantiger for this explanation.

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by sheelanadh » Tue Aug 10, 2010 3:05 am
clock60 wrote:x55, 8 numbers ( x is any digit exept 0 and 5. 8*1*1)
5x5, 9 numbers ( x any number exept 5. 1*9*1)
55x, 9 numbers ( x any number exept 5 1*1*9)
total=8+9+9=26
my pick for B

thank you........can u explain me little bit clear explination.....