Permutation Help required!!

Question

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 choice 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

fluke · Answer

Let the 3 digit number be of the form;
XYE

Possible digits for E: 0,2,4,6 i.e. Count=4
Possible digits for Y: 0,1,2,3,4,5,6 i.e. Count=7
Possible digits for X: 1,2,3,4,5,6 i.e. Count=6  
Total Possibilities = 4*7*6 = 168.

In the question, it is not mentioned that the digit can't be repeated.

Thus,
1,1,2
2,2,2
are all possible combinations.

*******************************************
If we were told that the digits can't be repeated;

Then,
Possible digits for E: 0,2,4,6
Let's consider two cases; Unit digits as 0 and Unit digit as non-zero
CaseI:
E can be 0
X can be {1,2,3,4,5,6} Count = 6
Y can be any 5 out of {1,2,3,4,5,6}. Count= 5
Total =6*5 = 30

CaseII:
E can be 2,4,6. Count=3
X can be any 5 of {1,2,3,4,5,6} Count = 5
Y can be any 5 out of {0,1,2,3,4,5,6}. Count= 5

Total = 5*5*3=75
Thus; Total possibilities = 30+75= 105

karthikp1211 · Answer

Case (i): 3 digit number ending with 0
 
 First 2 digits can be selected from (1,2,3,4,5,6) in 6p2 ways = 30

Case(ii): 3 digit number ending with (2,4 or 6)

units digit can be selected in 3 ways.(2,4,6)
 Tens digit can be selected in 6 ways.(0,1,2,3,4,5,6)
 Hundreds digit can be selected in 5 ways (1,2,3,4,5,6) 
 = 3*5*5 = 75
 Hence 30+75= 105

Correct me if im wrong...

jko · Answer

Biggest question here is whether digits can be re-used or not.

Andrusha · Answer

guys
why not to count how many ALL (even&odd) numbers of 3 digit and than half of that would be even?

so (7!/4!)/2 is 105

KarishmaB · Answer

Given digits 0-6, in how many ways can you make 3 digit numbers?

If you re-use the digits: 6*7*7 (which is the case in the question since it doesn't say that you cannot re-use)
If you do not re-use the digits: 6*6*5

How did you get 7*6*5?

Also number of even and odd numbers would be equal if you have equal even and odd digits. Here you have 4 even digits (0, 2, 4, 6) and 3 odd digits (1, 3, 5). Also 0 complicates the matter because the number cannot have 0 at hundred's place. So number of numbers ending in 0 is not the same as number of numbers ending in 1 or 2 or 3....

anuu · Answer

Thanks guys for the reply..
T
I came across this question in urch website. The question doesn't mention whether the digits can be re-used or not. I guess in the gmat exam, the problem wud clearly state with or without repetition.

Anuu

gmat1220 · Answer

I also think the order in which we choose the digit matters. Primarily we MUST choose the unit's digit first - its the most restricted digit. And then we must choose a non-zero hundred's digit and finally the ten's digit.

Permutation Help required!!

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