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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
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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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