How many 3 digit number exist having exactly two 5's in them ?
A. 24 B. 26 C. 27 D. 29 E. 30
Permutation Problem
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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
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
- indiantiger
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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)
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)
"Single Malt is better than Blended"
Thanks. indiantiger for this explanation.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)
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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.....