Permutation Problem

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

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

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)

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.

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.....