No. of four-digit even numbers formed by digits 0 1 2 3 4 5 without repetition

Question

I started from the units. There can only be $3$ possibilities $0,2,4$. Then there are $5$ possibilities for the tenth place. THen there are $4$ possibilities for the hundredth place. Then only $3$ possibilities remain. Thus the permutation is $3 \times 4 \times 5\times 3 = 180$. Is it correct? Need guidance.

Ok, but with a small correction : zero cannot be the first digit, so that has to be eliminated as a possibility, which is not done above as we are giving the last digit a free choice. So you have to be a little more careful. — Sarvesh Ravichandran Iyer, Mar 14 '18 at 10:17

score 3 · Accepted Answer · answered Mar 14 '18 at 10:31

4 digit even numbers(including $0$ at the beginning) = $3* 5* 4* 3 = 180$

for choosing the last digit we have 3 ways {0,2,4}. Now we are left with 5 numbers, for rest 3 places $5*4*3$ ways

Now we will subtract those 4 digit even numbers with $0$ as the first digit = $2 * 4 * 3= 24$.
First digit is fixed = 0, 1 way. Now we are left with 2 choices for last digit (2,4). For choosing last digit 2 ways.
Now we have 4 digits and 2 places, $4*3$ ways.

Req.number of numbers = $180 - 24 = 156$

score 0 · Answer 2 · answered Mar 17 '18 at 18:26

I have two other approaches to this question as well

Ap1/ We have 2 cases

Case 1: Fix last digit is 0 => 1 way to choose last digit. Now consider first 3 digit First digit cannot be 0 => 5 ways Second digit cannot be 0 and as same as the first digit => 4 ways Third digit cannot be 0 and as same as the first and the second digit=> 3 ways Altogether, we have 5x4x3x1 = 60 ways

Case 2: Last digit can be 2 or 4 => 2 ways. First digit cannot be 0 or cannot be as same as last digit => 4 ways Second digit cannot be as same as the first and the last digit => 4 ways Third digit cannot be as same as the first, the second and the last digit => 3 ways Altogether, we have 4x4x3x2 = 96 ways

From case 1 and 2, we have 156 ways

Ap2/ a) How many 4 digits numbers can be formed 0,1,2,3,4,5 without repetition? Clearly, there is 5x5x4x3 = 300 (not begin with 0)

b) How many 4 digits odd number can be formed by 0,1,2,3,4,5 without repetition? Last digit can be 1 or 3 or 5 => 3 ways First digit can be anything except 0 and the last digit=> 4 ways Second digit can be anything except being as same as the first and the last digit => 4 ways Third digit then 3 ways Altogether, we have 3x4x4x3 = 144

(a) - (b) = 300-144 = 156 ways

score -1 · Answer 3 · answered Nov 03 '18 at 01:40

case(1): If repetition is not allowed, 1’s place can be filled by the nos (0,2 or 4) in 3P1 ways = 3 ways 10’s, 100’s and 1000’s places can be filled by 5 x 4 x 3 = 60 ways Therefore, total number of arrangements = 60 x 3 = 180. If ‘0’ comes in the 1000’s place, the number becomes 3 digits. Such 3 digits numbers =1 x 3 x 4 x 2 = 24. Total number of 4 digit even numbers = 180  24 =156.

No. of four-digit even numbers formed by digits 0 1 2 3 4 5 without repetition

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