Permutation Help Required!!

[anuu]

How many even numbers of three digits can be formed with digits 0,1,2,3,4,5 and 6?
Here is my approcah:

The last(unit) digit has to be 0 , 2 , 4 or 6"
So there woeld be 4 cases:

CASE 1:

With '0' as the last(unit digit):

there r 4 choices for the last digit(0,2 , 4 or6) and we select '0',
so there r 6 digits remaining.

there first 2 digits can be selected from the remaining 6 digits in 6p2
ways i.e 30 ways.

So total number if ways for forming 3 digit even number ending with '0' is
4*30 = 120 ways.

CASE 2:

there r 4 choice for the last digit(0,2 , 4 or6) and we select say '2',
so there r 6 digits remaining.

Now, the 10th digit(middle digit), has to be '0' , so there is only one way
for filling the 10th digit. So there r 5 digits remaining

The 100th digit (first digit) can be selected from the reamining 5 digits in
5p1 ways i.e 5 ways

So total number if ways for forming 3 digit even number ending with '2' is
4*5 = 20 ways.

For Case 3 and Case 4: the same logic as Case 2 applies. Since hte 10 digit is '0'.
So for case 3 and 4 , So total number if ways for forming 3 digit even number is 20.

Total number of ways is thus: 120+20+20+20 = 180 ways

Is this approach/answer rite?

The question was posted in a website and there were many answere(120, 105, 168 etc) not sure which is the rite answer?

Anu

[manpsingh87]

How many even numbers of three digits can be formed with digits 0,1,2,3,4,5 and 6?
Here is my approcah:

The last(unit) digit has to be 0 , 2 , 4 or 6"
So there woeld be 4 cases:

CASE 1:

With '0' as the last(unit digit):

there r 4 choices for the last digit(0,2 , 4 or6) and we select '0',
so there r 6 digits remaining.

there first 2 digits can be selected from the remaining 6 digits in 6p2
ways i.e 30 ways.

So total number if ways for forming 3 digit even number ending with '0' is
4*30 = 120 ways.

CASE 2:

there r 4 choice for the last digit(0,2 , 4 or6) and we select say '2',
so there r 6 digits remaining.

Now, the 10th digit(middle digit), has to be '0' , so there is only one way
for filling the 10th digit. So there r 5 digits remaining

The 100th digit (first digit) can be selected from the reamining 5 digits in
5p1 ways i.e 5 ways

So total number if ways for forming 3 digit even number ending with '2' is
4*5 = 20 ways.

For Case 3 and Case 4: the same logic as Case 2 applies. Since hte 10 digit is '0'.
So for case 3 and 4 , So total number if ways for forming 3 digit even number is 20.

Total number of ways is thus: 120+20+20+20 = 180 ways

Is this approach/answer rite?

The question was posted in a website and there were many answere(120, 105, 168 etc) not sure which is the rite answer?

Anu

here only two cases are possible:
case 1) when '0' is at the unit place.
case 2) when '0' is not at the unit place.

case 1) when 0 is at the unit place - - 0; then the digit at the hundred's place can be filled in 6 ways and the digit at the ten's place can be filled in 5 ways, hence total no. of such numbers = 6*5*1=30;

case 2) when 0 is not at the unit place, then unit place can be filled by any of the 2,4,6 i.e. there are 3 ways for filling the unit digit, now as zero is not fixed therefore the total no. of ways for filling the hundred's place is 5(because if we consider 6 than it will also include those cases when 0 will be at the hundred's position in that case number would become a two digit number), now as two places are occupied, the remaining ten's place can be filled up by any of the remaining five numbers in 5 ways. hence total no. of such numbers would be 3*5*5=75

hence total no. of numbers= case1+case2= 30+75=105;

[virendersingh1]

Nothing is mentioned abt Repetition. Taking repetition by default as allowed .

Unit Place: 0,2,4,6
Hundreds Place: 1,2,3,4,5,6
Tens Place: 0,1,2,3,4,5,6

Total=4*6*7=168

Without Repetition:
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Same as manpsingh87 did.