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Q.How many four-digit numbers, with distinct digits are there such that the sum of the digits of each of these numbers is an odd natural number?
a.2160  b.2090  c.1880  d.2376

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

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