How many 3-digit positive integers are odd and do not contain the digit 5?

Solution:

For an odd number, unit place has to be odd.

CASE 1:

 If repetition of digits allowed, excluding 5

hundred

tenth

unit

→ Fill the 100th place by any of the digits from 0 to 9, excluding 0 and 5, this can be done in 8 ways.→ First fill the unit place by any of the odd numbers 1, 3, 7, 9 (∵ 5 is excluded).

This can be done in 4 ways.

→ Then fill the 10th place by any of the digits from 0 to 9, excluding 5.

This can be done in 9 ways.

Number of odd number possible = 8 × 9 × 4 = 288 [by fundamental principle of counting]

CASE 2:

If repetition of digits is not allowed and excluding 5.

→ Since the 3 digit number is odd, let us first fill the unit place by an odd number, say 1.

→ Then fill the 100th place by any of (0 to 9), excluding 1 and 5, this can be done in 7 days.

Number of odd numbers having 1 unit place = 7 × 7 × 1 = 49.

This process is repeated by having 3, 7, 9 in unit place.

Number of 3 digit odd numbers possible without 5 = 4 × 49 = 196.

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How many 3-digit positive integers are odd and do not contain the digit 5?

Summary:

For 3-digit positive integers are odd and do not contain the digit 5, CASE 1: If repetition of digits allowed, excluding 5 ⇒ Number of odd number possible = 8 × 9 × 4 = 288 CASE 2: If repetition of digits is not allowed and excluding 5 ⇒ Number of 3 digit odd numbers possible without 5 = 4 × 49 = 196