800guy 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?

A very crude approach.. please give the simpler way
(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

3 digit numbers

first digit can be chosen in 5 ways excluding 0, for each of these 5 ways second digit can be chosen in 5 ways excluding the first number and including 0, third digit can be chosen in 4 ways excluding the first 2 digits.

5 x 5 x 4 = 100

# odd => last digit should be 1, 3, 5

last digit can be chosen in 3 ways, for each of these first digit can be chosen in 4 ways excluding 0 and the second digit can be chosen in 4 ways (including 0)

4 x 4 x 3 = 48

# even = total - # odd = 100 - 48 = 52

kulksnikhil wrote:

800guy 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?

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

After you found out the total and # even why did work out for odd again. It is just the total - # even. This would have probably saved some time.

Check this website

https://www.mansw.nsw.edu.au/members/ref ... no4yen.htm

and check the sticky "two great guides on counting methods and probability" in the GMAT Math forum.

OA:

Odd: fix last as odd, 3 ways __ __ _3_
now, left are 5, but again leaving 0, 4 for 1st digit & again 4 for 2nd digit: _4_ _4_ _3_ =48 Odd.

100-48= 52 Even

Great explanations!

There is a direct way also for calculating the no. of even numbers


If 0 is selected as the units digit
Then the no. of combinations = 5 * 4 * 1 = 20

If 0 is not selected as the units digit then the no. of combinations = 4 * 4 * 2 = 32

Hence total number of even no.s = 30 + 20 = 52

total numbers = 5*5 (considering 0 too) * 4 = 100

odd digits = units place = (1,3,5) = 3
tens and hundred's digit = 4*4

thus total = 4*4*3 = 48
even = 100-48 = 52