in Combinatory edited by
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How many $4$-digit even numbers have all $4$ digits distinct?

  1. $2240$
  2. $2296$
  3. $2620$
  4. $4536$
in Combinatory edited by
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4 Comments

Same question could be done in one more way.

(All four digit no.) - (All four digit no. that are not even).
1
can you write whole ans  for your approach  ?
1
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
0
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

source: below answer
1

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8 Answers

58 votes
 
Best answer
  • 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 by

21 Comments

edited by
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.
0

@Verma Ashish

plz read  comment carefully.

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

0

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..

0

@Verma Ashish

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

0

@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. 

2

@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.

0

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

1

@srestha 

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..

1

@Verma Ashish

@srestha

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

See this 

1

@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.

0
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..
1

@srestha

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

0
which one?? I saw both following orders.
0

 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

0
  • 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

@Verma Ashish asked it.

 

0

@srestha ma'am

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

1
it is maintaining for both.
0

@srestha

No, ma'am  

it is maintaining for both.

see this for case 1

Why this gives wrong answer for case1?

0

@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??

0
Okay
0
8 votes
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 by
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
0
6 votes
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

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
0
5 votes

$Method\ 1:$

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

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


$Method\ 2:$

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

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

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

1 comment

great one
0
0 votes

Simple solution for this problem

0 votes
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
0 votes
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
0 votes
2296 is the correct answer.
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