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P and C hard question

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by chaitanyareddy » Tue Nov 09, 2010 8:22 pm
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 answer.
Even the word Impossible says I'Mpossible.
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by diebeatsthegmat » Tue Nov 09, 2010 8:31 pm
chaitanyareddy 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 answer.
ABC is the 3 digit number we must find and the digits arnt allowed to repeated....
A cant be 0 thus A has 5 choices
B =6-1=5 ways to choose from (0,1,2,3,4,5) and we have to - 1 because A was chosen
the same to C=4
total= 5*5*4=100?

even number ABC
C must be even thus C= 3 choices (0,2,4)
A cant be 0 and cant be any number of C thus A can have only 4
C= 4
so total=48?
the same with ABC -odd
whats the answer
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by chaitanyareddy » Tue Nov 09, 2010 10:57 pm
Yeah even I am getting the same answer. However one doubt is if there are total 100 possible 3 digit numbers and 48 of them are even and 48 of them are odd , then what about the remaining 4 numbers...?

I mean all the three digits numbers able to be formed should be equal to number three digit even + no of three digit odd.

but 100 ≠ 48 + 48.

Am I missing something here..?
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by vijaynaik » Tue Nov 09, 2010 11:40 pm
Even#s are 52 and ODD#s are 48.

Since u know the ODS#s let me give u how the EVEN number count comes to 52.

Case 1: Number ending with 0. 5*4 (1st digit can not be 0 so u have 1,2,3,4,5 to select from. 2nd digit can not be 0 and the 1st digit. So u have 4 numbers to select from).
Case 2: Number ending with 2. 4*4
Case 3: Number ending with 4. 4*4

Total = 52.
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