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Korolek [52]
2 years ago
15

How many 4-digit positive integers are there for which there are no repeated digits, or for which there may be repeated digits,

but all digits are odd?
Mathematics
1 answer:
BARSIC [14]2 years ago
7 0

The number of ways of arranging 4-digit positive integers with no repeated digits is 4536 ways and number of ways of 4-digit positive integers with repeated digits, but all digits are odd is 625 ways.

In this question,

Positive integers are 0,1,2,3,4,5,6,7,8,9

Total number of integers = 10

This can be solved by permutation concepts.

Case 1: 4-digit positive integers with no repeated digits,

First digit, cannot be zero. So remaining 9 digits.

Second digit, can be any digit other than the first digit. So 9 digits.

Third digit, can be any digits other than first and second. So 8 digits.

Fourth digit, can be any digits other than first, second, third digit. So 7 digits.

Thus, Number of ways of 4-digit positive integers with no repeated digits ⇒ (9)(9)(8)(7)

⇒ 4536 ways.

Case 2:  4-digit positive integers, there may be repeated digits, but all digits are odd

Odd integers are 1,3,5,7,9

Number of digits = 5

In this case, we can repeat the digits. So all places can have 5 possibilities.

Thus number of ways of 4-digit positive integers with repeated digits, but all digits are odd = (5)(5)(5)(5)

⇒ 625 ways.

Hence we can conclude that the number of ways of arranging 4-digit positive integers with no repeated digits is 4536 ways and number of ways of 4-digit positive integers with repeated digits, but all digits are odd is 625 ways.

Learn more about permutation here

brainly.com/question/18650638

#SPJ4

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