How many 4 digit numbers with distinct digits such that sum of digits is odd?

Efficient Approach:  We have to fill (n) places with different digit. Like we n=2 then (_ _) places to fill. first place we fill (1 to 9) any number. let we fill 9 in first place then in second place we have choice (0 to 8). So for first place we 9 choices because we can not fill 0 at first place and after that for 2nd place we 9 choice and for 3rd place we 8 choice then so on… 

Hint: We have to find the four-digit number. First, take the ten’s place and assume how many odd numbers can be placed in it, then go to thousands of places and assume the number of digits can be placed in it.

Complete step-by-step answer:
The number should be formed with a number of digits is 4.
It should be noted that the four digits odd number that we have to form should be less than 10000 (the last 4 digit number) and greater than 1000 ( the last 3 digit number)
Now, the four digit number consists of one’s, ten’s, hundreds and thousands digits, so the possible one’s digit will be 1, 3, 5, 7 and 9 because we have to form the odd number, so 1, 3, 5, 7and 9 are odd numbers it means one of the total 5 numbers can be placed at one’s digit.
Then the thousand’s place can be one of the 1, 2, 3, 4, 5, 6, 7, 8, and 9, but one of them is already occupied by one’s digit, then only one of the 8 numbers is possible to place in the thousand's number.
And the hundred’s also the same, 8 possible numbers, and coming to the ten’s place, we will have only 7 possible numbers to place in it.
Then the number could be \[8 \times 8 \times 7 \times 5 = 2240\]
So the number of four-digit odd numbers can be formed is 2240. It means option (A) is correct.

Note: Here, we have to know the definition of the odd number and the probability. We have to come through how a number can be formed with the help of the probability, while concluding the digit in one’s place, we have to take only odd numbers.

Even number set { 0, 2, 4, 6, 8 }   Odd number set  { 1, 3, 5, 7, 9 } 

The answer can be achieved by considering 2 cases

1. 3 digits are even and 1 digit is odd

2. 3 digits are odd and 1 digit is even

1st case: 

  3 digits are even and 1 digit is odd 

      here 2 cases are possible     

• 2 non-zero even digits and 1 digit = 0 and 1 odd digit​​​​​​

      OR

• 3 non-zero even digit and 1 odd digits ​​​​​​​

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• 2 non-zero even digits and 1 digit = 0 and 1 odd digit​​​​​​

           Choosing 2 non-zero even digit = 4C2 = 6

           Choosing 1 odd digit = 5C1 = 5

           Arranging all 4 digits = 3 × 3 × 2 × 1 = 18

           The no. of ways will be = 18 * 6 * 5 = 540    [the numbers can't be started with 0] 

• 3 non-zero even digit​​​​​​​ and ​​​​​​​1 odd digits ​​​​​​​

           Choosing 3 non-zero even digit = 4C3 = 4

           Choosing 1 odd digit = 5C1 = 5

           Arranging all 4 digits = 4! = 24

           The no. of ways will be = 24 * 5 * 4 = 480

The total no. of ways of 1st case = 540 + 480 = 1020

Now the 2nd case

3 digits are odd and 1 digit is even

• 1 digit = 0​​​​​​​ and 3 odd digits 

       OR

• 1 non-zero even digit​​​​​​​ and 3 odd digits 

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

• 1 digit = 0​​​​​​​ and 3 odd digits 

           Choosing 3 odd digits = 5C3 = 10
           Arranging all 4 digits = 3 × 3 × 2 × 1 = 18

           The no. of ways will be = 18 * 10 = 180 

• 1 non-zero even digit​​​​​​​ and 3 odd digits 

           Choosing 1 non-zero even digit = 4C1 = 4

           Choosing 3 odd digit = 5C3 = 10
           Arranging all 4 digits = 4! = 24

           The no. of ways will be = 24 * 4 * 10 = 960

The total no. of ways of 2nd case = 180 + 960 = 1140

The total no. of ways of 1st case + The total no. of ways of 2nd case = 1020 + 1140 = 2160 ways

How many 4 digit numbers are there with distinct digits with each digit odd?

How many four

How many 4 digit numbers can be made up of only odd digits repetitions?

How many 4 digit numbers have distinct digits?