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

  1. 2240
  2. 2296
  3. 2620
  4. 4536
asked in Combinatory by Veteran (59.4k points)
retagged by | 2.3k views
0
Same question could be done in one more way.

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

2 Answers

+35 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 1×9×8×7=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 4×8×8×7=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 (327 points)
selected by
0

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 ?

+1
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.
0

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 ?

0

@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. 

 

+1
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??
0

 Puja Mishra yes it is wrong.The number can't start with 0.for example 0372 should not be accepted.

0
i havnt considered 0 fr 1st place ....
0
then explain choices at each position
0
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 ??
+4
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.
0
sorry i hav made a mistake during typing it ... nw see ..
+1
I didn't notice typing mistake.I understood what you want to say.see my comment again and read it carefully.
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 ...
+2
@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..
+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
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