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Permutation problem.

by vertigo05 » Sun Apr 04, 2010 6:21 am
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?

I don't know the exact source of this problem. I found this in a quant problems collection. The answers given are 48 odd and 52 even.
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by sumanr84 » Sun Apr 04, 2010 7:25 am
vertigo05 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?

I don't know the exact source of this problem. I found this in a quant problems collection. The answers given are 48 odd and 52 even.
Total number of 3 digit number that are possible are = 5 * 5 * 4 = 100 ( first place can be taken by 1,2,3,4,5 but not 0 )

No of even number will be composed of numbers ending with 0, 2 and 4.

when 0 is the last digit, we have 5 options for 1st place and 4 options for 2nd place ==> total = 5*4 = 20

when 2 is the last digit, we have only 4 options for 1st place and 4 options for 2 place ==> total = 4*4 = 16 ( 0 cannot be at the first place )

similarly, for 4 also we have 16 numbers.

so, total number of even numbers are = 20+16+16 = 52. Hence, odd number = 48
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by this_time_i_will » Tue Apr 06, 2010 6:58 pm
vertigo05 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?

I don't know the exact source of this problem. I found this in a quant problems collection. The answers given are 48 odd and 52 even.
i was just wondering, if you start filling from units place then we have following:
total number of 3 digits = 6*5*3 = 90.
however, if we start from hunderds place, we have following:
5*5*4 =100...
can any one please clear my confusion
Last edited by this_time_i_will on Wed Apr 07, 2010 5:23 pm, edited 1 time in total.
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by gmat.cracker24 » Wed Apr 07, 2010 4:46 am
Total no. of 3 digits=(# of ways of filling hundredths place)X(# of ways of filling tenth place)X(# of ways of filling unit place)
=5(Cant have zero)X5(remainig 5 digits)X(Remaining 4 digits)
=100
No. of even digits=(Cant use twodigits out of 2,4,0 since they are used in making it either even or three digits)X(Remainig digits)X (Either2,4,0)
=4X4X3
=48
Odd digits=100-48=52
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