A very crude approach.. please give the simpler way800guy wrote:How many number of 3 digit numbers can be formed with the digits 0,1,2,3,4,5 if no digit is repeated in any number? How many of these are even and how many odd?
(Please do not read if you don't have time)
I> Total number of three digit numbers
since 3 have to be selected.
6P3 would give the number of combinations (Without number repetition)
= 6! / 3! = 120
Since there are 6 numbers altogether... every number will remain in one position with in the 3 digits for 120/6 times so zero will be in first place for 120/6 = 20 times
taking out numbers starting with zeros = 120 -20 = 100 3 digit numbers can be formed.
"A total of 100 3 digit numbers can be formed."
II> Number of Even and Odd numbers
With in 20 numbers starting with zero, 5 digits can take last place (1~5, excluding zero b'coz no repetition of number)
So number of numbers ending with same number, starting with zero = 20/5 = 4
there are two even number between 1~5 i.e. 2 and 4
there are three odd numbers between 1~5 i.e 1,3 and 5
II> Number of even numbers =
Out of 120, number of numbers ending with 0,2,4 in last place
= 20*3 = 60
Number of numbers with zero in first place and 2 or 4 in last place = 2 *4 = 8
total number of 3 digit even numbers = 60 - 8 = 52
III> Number of Odd numbers =
Out of 120, number of numbers ending with 1,3,5 in last place
= 20*3 = 60
Number of numbers with zero in first place and 1,3,5 in last place = 3 *4 = 12
Total number of 3 digit odd numbers = 60 - 12 = 48
To summarise:
(took me more than 15 minutes to solve.. not at all a GMAT approach)
Number of 3 digit numbers: 100
Number of even 3 digitis : 52
Number of odd 3 digits : 48
