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

1. $2240$
2. $2296$
3. $2620$
4. $4536$
edited | 2.7k views
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Same question could be done in one more way.

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

• 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.

edited by
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Suppose if count all the 4 digit numbers not starting with 0, I would have 9 x 9 x 8 x 7 = 4536 (9 choices for first digit excluding 0, 9 choices for the second digit excluding the first chosen digit, followed by 8 and 7 for the other two digits)

Out of 4536 half should be odd and the other half should be even, so 4536/2 = 2268 odd and 2268 even numbers

But my answer is not matching with the above solution, where have I gone wrong ?

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Issue is '0'. Since it is not allowed as the first digit, we are forced to take one among 1-9. Now, this makes the selected digit not being among the unit digit. We have 5 odd digits - 1, 3, 5, 7, 9 which can be eliminated like this but only 4 even digits can be eliminated 2, 4, 6, 8. Thus among all the possible numbers, even numbers will be more than odd numbers.
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I could not understand what you meant by Now, this makes the selected digit not being among the unit digit. Can you please tell some numbers which are missed out ?

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@Registered user,

"Out of 4536 half should be odd and the other half should be even, so 4536/2 = 2268 odd and 2268 even numbers"

How did u conclude that there are equal amount of odd and even numbers? Those 4536 numbers are not consecutive. They are numbers having DISTINCT digits.

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4 digits ... 4th place can be filled in 5 ways {0,2,4,6,8}

1st place can be filled in 8 ways

2nd place can be filled in 8 ways

3rd place can be filled in 7 ways ...

so ans is 8*8*7*5 = 2240 ...  is it wrong??
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Puja Mishra yes it is wrong.The number can't start with 0.for example 0372 should not be accepted.

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i havnt considered 0 fr 1st place ....
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then explain choices at each position
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for 4th place {0,2,4,6,8} which is 5

for 1st place all numbers except 0 and digit in 4th place which is 8

for 2nd place all numbers except digit in 1st and last place which is 8

for 3rd place all numbers except digit in 1st,2nd and last place which is 7

is it correct ??
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suppose you choose the number '0' for 4th place.Now according to you for 1st place, there are 8 choices(for 1st place all numbers except 0 and digit in 4th place ) but we already choose number 0 4th place and according to question number should be distinct. you are subtracting 1 extra choice for this particular case.
Check the given solution.Now you can able to understand your mistake.
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sorry i hav made a mistake during typing it ... nw see ..
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I didn't notice typing mistake.I understood what you want to say.see my comment again and read it carefully.
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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 ...
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@Puja As Leen told, suppose your 4th digit is 0, then the first digit has 9 possibilities and not 8.
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@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

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Ok got it... i hav to consider two cases ...
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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??
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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.

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could you please explain it more? thankyou
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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
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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
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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