+20 votes
3.1k views

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

1. $2240$
2. $2296$
3. $2620$
4. $4536$
asked
edited | 3.1k views
0
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  ?

## 3 Answers

+42 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.

answered by (337 points)
edited by
0
Still i am nt able to understand my mistake ... i hav done this ...i am nt considering 0 at unit place ... all i am doing combining the cases ... i hav {0,2,4,6,8} for 4th place ... nw 1st place cant be 0 and digit at 4th ... so 8 choices ... fr 2nd 8 choices .. as we cant chose 1st and 4th place digit.. fr 3rd place 7 choice as digits fr 1st 2nd 4th cant be selected ...
+3
@Puja As Leen told, suppose your 4th digit is 0, then the first digit has 9 possibilities and not 8.
0

@Puja Mishra

It may be your mistake :-

suppose at 4th position you selected 0

then for the first position you have 9 choice (1-9) and not 8 choices

+1
Ok got it... i hav to consider two cases ...
0
considering 2nd case "If the number is even ending with something else than 0 then there are 4 choices for the last digit, 9 choices for the second last digit , 8 for the third last digit and 6 for the first digit as 1st can't be 0,

what am i doing wrong here??
0
suppose you choose number 0 for 3rd place or 2nd place but you are also subtracting the case when the 1st number is 0.In that case, your answer will be wrong.

Follow the answer given by Arjun sir.
0
could you please explain it more? thankyou
0
read all comments..
0

I agree with $1st$ case of the selected answer. However, I am getting a different result for the $2nd$ case. Where I am going wrong?

My understanding:

• There are $4$ choices for the units digit (either of $2, 4, 6, or \ 8$).
• $9$ choices for the tens digit.
• $8$ choices for the hundredths digit.
• $6$ choices for the thousandths digit (no $0$ and none of the digits selected for the previous positions)

$\implies 6 \times 8 \times 9 \times 4 = 1728$ options.

0

We can deal with $'0'$separately.

+7 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
answered by Active (1.5k points)
0
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
+3 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
answered by Active (3.9k points)
0
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
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