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3-digit numbers


Ashie

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Thanks Bullfighter. Yep, the answer is what you have given.

 

A small clarification - is it that we have to always find the number of Odd first, then do 100-48 to get the even? If we have to calculate even first, how do we do that? I know it's easier to find odd and then subtract from 100 to get even. But still would like to know how to go about the other way.

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you can find either way, But, finding # even is not as simple as finding # odd. Here, we have to consider Two diff. cases. One when 0 is used at the last digit, & hence automatically can not be used at the 1st place. & the other one where 0 is Not used at the last place...

 

Even: For last digit: 0, 2, 4

 

1. Last digit: 2 & 4. Other left: 0,1,3,5.

 

4*4*2= 32

 

2. Last as 0. Other left: 1,2,3,4,5.

 

5*4*1= 20

 

Total= 52.

 

 

Odd= 100-52= 48.

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for the original Q, 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?

 

Choices available:0,1,2,3,4,5. ie. Six in total.

 

1000'th digit can be filled in 5 ways (as 0 not allowed), now, repetition not allowed, hence left ar 5 digits, Hence, 100th digit can also be filled in 5 ways. Again, left are 4, hence Unit's digit can be filled in 4 ways.

 

Hence, total ways: 5*5*4 (Permutation)

 

___________________________

 

for Odd. Unit's digit can be filled in 3 ways. (1,3,5 As Odd)

Left are 5 digits, again as 0 can not be used, for 1000th digit choices left are 4. (one used as Unit digit & also excluding 0, 6-2=4)

& for 100th digit, 2 already used, one at unit's place & one at 1000th place, laft are 4 choices.

 

Hence, 4*4*3

 

Similarly for Even, as shown above.

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you can find either way, But, finding # even is not as simple as finding # odd. Here, we have to consider Two diff. cases. One when 0 is used at the last digit, & hence automatically can not be used at the 1st place. & the other one where 0 is Not used at the last place...

 

Even: For last digit: 0, 2, 4

 

1. Last digit: 2 & 4. Other left: 0,1,3,5.

 

4*4*2= 32

 

2. Last as 0. Other left: 1,2,3,4,5.

 

5*4*1= 20

 

Total= 52.

 

 

Odd= 100-52= 48.

Thanks a bunch Bullfighter!

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Hi bullfighter,

 

Thanks for ur explanation. However, Can u pls tell me why did u split the method of finding EVEN numbers into two parts. One where 0 is at the end & the other where it is not?

 

Y can we not consider it as one option - If it has to be even then the last digit should be any of the three 0,2,4.

so there are 3 possible ways for third digit,

4 for first

4 for second

what is wrong in this method?

I know the ans does not match ,it is just that I want to figure out what is missed in this approach

Can u pls help me out

thanks

adroja

you can find either way, But, finding # even is not as simple as finding # odd. Here, we have to consider Two diff. cases. One when 0 is used at the last digit, & hence automatically can not be used at the 1st place. & the other one where 0 is Not used at the last place...

 

Even: For last digit: 0, 2, 4

 

1. Last digit: 2 & 4. Other left: 0,1,3,5.

 

4*4*2= 32

 

2. Last as 0. Other left: 1,2,3,4,5.

 

5*4*1= 20

 

Total= 52.

 

 

Odd= 100-52= 48.

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