7,344 views

How many $4$-digit even numbers have all $4$ digits distinct?

1. $2240$
2. $2296$
3. $2620$
4. $4536$

Same question could be done in one more way.

(All four digit no.) - (All four digit no. that are not even).
can you write whole ans  for your approach  ?
Case 1: Pos1 is for odd digit:

pos1 : 5 option

pos4 : 5 option (even number including 0)

pos2 : 8 option (two numbers already used, no repetition allowed)

pos3 : 7 option

total =

case 2: pos 2 is for even digit

pos1 : 4 option (0 cannot be there as its 4 digit number)

pos4: 4 option (one of 2,4,6,8 already used so 3 left + we have 0 so 4 options)

pos2 : 8 option

pos1 : 7 option

total = 5*5*8*7 + 4*4*8*7 = 1400 + 896 = 2296
I have tried in this way -

No of 4 digit even numbers= Total no of 4 digit even numbers -  Number having 0 at the front

Total no of 4 digit even numbers =last place is filled in 2,4,6,8,0 ie 5 ways. Next place in 9 ways, next in 8 ways and next in 7 ways (go from right)

Number having 0 at the front = last place is filled in 2,4,6,8 ie 4 ways, next place in 8 ways and next in 7 ways and first place is for 0.

Required number of numbers = (7*8*9*5)-(7*8*4)  = 2520-224 = 2296

### Subscribe to GO Classes for GATE CSE 2022

• If the number ends with a $0$ then there are $9$ choices for the first digit, $8$ for the second and $7$ for the third, which makes $1\times9\times8\times7= 504$ possibilities.

• If the number is even ending with something else than $0$ then there are $4$ choices for the last digit, $8$ choices for the first digit (no $0$ nor the last digit), $8$ for the second digit and $7$ for the third digit, which makes $4\times8\times8\times7 =1792$

Together, this gives $2296$ numbers with $4$ distinct digits that are even. Note that this does not allow leading $0$, as you see to want it based from the question.

Correct Answer: $B$

edited
In your first case like $3012,3102,5034,5304.....$, numbers are missing. We know that we can't place $0$ in the first place. But you completely ignore the $0$, which lead to an incorrect answer.

Hope you understand.

@Verma Ashish

how do u write a number?? from left side or from right side??

Hello sir,

In your first case like 3012,3102,5034,5304.....

No these are not missing..

For 3012-

2 in the last(4th) place.

Now there are 9 choices for 3rd place so 1 is possible at 3rd place.

8 choices remaining  for 2nd place so 0 is possible.

For 1st place 6 choices remaining 3 can be there..

Similarly 3102 is also possible..

@Verma Ashish

how do u write a number?? from left side or from right side??

@Verma Ashish

Brother i'm not sir

Please call me brother or my name.

No these are not missing..

For 3012-

$2$ in the last(4th) place.

Now there are 9 choices for 3rd place so $1$ is possible at 3rd place.

8 choices remaining  for 2nd place so $0$ is possible.

For 1st place 6 choices remaining 3 can be there..

Similarly 3102 is also possible..

When you choose $0$ at 2nd or 3rd place then 1st place will have $7$ choices not $6$ choices.

@srestha ma'am

how do u write a number?? from left side or from right side??

I satisfy the even number condition(last digit should be $0\:or\:2\:or\:or\:4\:or\:6\:or\:8$). After that order doesn't matter.

Order matter here. Otherwise @Verma Ashish calculation has no mistake I think.

eg:- how many 4 digit numbers possible with digits 1,2,3,4 (repitition not allowed)

4*3*2*1=4!

1*2*3*4=4!   Here left to right or right to left doesn't matter..

But i think in case of restrictions it matters..

@Verma Ashish

@srestha

Yes, order matters in case of restriction. Otherwise, it gives the wrong answer.

See this

@Verma Ashish

never we do , last place fill up first. Still if I think u r correct, though it is not a concrete contradiction of this question.

I mean last place fill up first  or first place fill up last, which one occuring not certain here.

Yes.

Even in case 2(as mentioned in pic in above one comment) of this question, order doesn't matter..

But from left to right is better..

@srestha

From my answer, Case$1$ follows the order and case$2$ not followed why?

which one?? I saw both following orders.

which one?? I saw both following orders.

What order follows here?

From the very 1st image which i added. For case$2$ we can write like this

4th place ⟹1 choice

3rd place ⟹9choices

2nd place ⟹8choices

1st place ⟹7choices

• 3rd place ⟹9choices
• 2nd place ⟹8choices
• 1st place ⟹7choices

just reverse them, order will follow,like

• 1st place ⟹9choices
• 2nd place ⟹8choices
• 3rd place ⟹7choices

@srestha ma'am

It is valid for case2 only,why not for case1?

it is maintaining for both.

@srestha

No, ma'am

it is maintaining for both.

see this for case 1

Why this gives wrong answer for case1?

@Lakshman Patel RJIT

try to understand , and read comments again before commenting and think some time.

I mean there that,  do both in ordered.

Is there any case (like case 1 and case 2 as u say) giving any error??

No. right??

But if u do unordered , u got error in some cases.

So, it is better to avoid unordered.

right??

Okay
Available digits are { 0,1 , 2  ... 9}

Even { 0 2 4 6 8}

Odd { 1 3 5 7 9 }

Pos 4 must be one of even number . But Pos 1 can be odd or even but not 0.

(case 1 )

POS 1: ODD { ONE OF 1 3 5 7 9}

POS 2:7C1

POS:3 8C1                                                 SO Total is : 5 X (7x8x5)=1400

POS 4: ONE OF EVEN NO i.e. 5C1

(CASE 2: POS1 IS EVEN INLUDING 0)

POS 1: EVEN { ONE OF 0,2,4,6,8}

POS 2:7C1

POS:3 8C1                                                 SO Total is : 5 X (7x8x4)=1120

POS 4: ONE OF EVEN NO EXCLUDING THE DIGIT OF POS1 i.e. 4C1

(Case 3: POS1 IS 0)

POS 1: 0

POS 2:7C1

POS:3 8C1                                                 SO Total is : 1 X (7x8x4)=224

POS 4: ONE OF EVEN NO EXCLUDING 0  i.e. 4C1

The ans is : case1 + case2 - case3 =(1400+1120-224)=2296

### 1 comment

reshown
Hi sir, I can't understood in this method .

Why we take 3cases  sir your answer right but  I don't understand the case 1 .I agree with puja dii ans
I have tried in this way -

No of 4 digit even numbers= Total no of 4 digit even numbers -  Number having 0 at the front

Total no of 4 digit even numbers =last place is filled in 2,4,6,8,0 ie 5 ways. Next place in 9 ways, next in 8 ways and next in 7 ways (go from right)

Number having 0 at the front = last place is filled in 2,4,6,8 ie 4 ways, next place in 8 ways and next in 7 ways and first place is for 0.

Required number of numbers = (7*8*9*5)-(7*8*4)  = 2520-224 = 2296
by

### 1 comment

Last digit must be something among 0,2,4,6,8.

1 point u must note here is that it is 1st choosing and then araanging .

Now there is 2 possibility

1st 1-  if 0 is at last position , if so,

Then the other 3 positions can be filled in 9p3 ways that is equal to 9*8*7

Now here comes the 2nd case

Here take care that 1st digit is not zero otherwise it won't make 4 digits .

We have already considered the case where last digit is 0 so now in this case last digit can be filled in 4 ways possibilities may be 2,4,6,8 means total 4

2nd and 3rd digit can b filled in 8p2 ways .

And first digit in 8 ways

So it becomes 4*8*8*7

So total it becomes 4*8*8*7+9*8*7

$Method\ 1:$

 $-$ $-$ $-$ $0$ $9.8.7.1$ $-$ $-$ $-$ $2$ $8.8.7.1$ $-$ $-$ $-$ $4$ $8.8.7.1$ $-$ $-$ $-$ $6$ $8.8.7.1$ $-$ $-$ $-$ $8$ $8.8.7.1$

$9.8.7.1+(8.8.7.1)\times 4=2296$

$Method\ 2:$

$1)\ Total\ \#\ of\ 4\ digit\ numbers\ and\ all\ distinct=9.9.8.7=4536$

$2)\ \#\ of\ 4\ digit\ odd\ numbers\ and\ all\ distinct=8.8.7.5=2240$

$Even=Total-Odd=4536-2240=2296$

### 1 comment

great one

Simple solution for this problem

by
Here in given question  we are find out 4 digit  distinct  number from the  decimal digit  0,1,2,3,4,5,6,7,8,9,

here see in above digit sequence  0 is included then we deal with  0 in deifferent manner if   they involved in  number

first we find those number in which 0 is not involved

------                                  ------                                                   ------                          -------

6                                         7                                                     8 {remaining}   4{2,4,6,8 possible}

now deal  with zero number

---------                  ----------                -----------           ----------------

7                          8                                9                     0{ here zero placed }

------------             -------------            --------------             ----------------

7                           8                     0 {placed }                     4

------------                    --------                  ---------            -----------

7                        0                                 8                        4

---------                    ------------             ------------            ---------------

0 not possible

if msb is  0 then number is not four digit

so total number =6*7*8*4+7*8*9+7*8*4+7*8*4=2296
Here 0 creates trouble for us

So first let us ignore it

WE have 9 numbers {1,2,3,4,5,6,7,8,9}

_ _ _ _ fill these 4 digits in order to get an even number : 9*8*7*4 we all know this one

Now we will only count the cases where 0 is present

_ _ _ 0  example 1420 this is a valid number, 8250 this is also valid

so we are clearly able to see there are a lot of such numbers which are 9*8*7 --------(1)

now fix 0 in any of the remaining 2 places all of those will give same answer  X _ _ X ( as u cant put 0 in MSB)

like imagine a number 8024 this is a valid number (remember it has to be even also)

so 2 cases, hereby multiplying with 2 as the number is of the form

NUM | 0 | NUM | EVEN

8*7*4 *2

sum all the cases

we will get 2296
by