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Q: How many different three digit numbers can be formed from the digits 0 through 9 if the first digit must be odd and the last digit must be even?

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If the digits are all different then 18. Otherwise, 192.

12 numbers.

Assuming you mean permutations of three digits, then the set of numbers that can be made with these digits is: 345 354 435 453 534 543 There are six possible permutations of three numbers.

Assuming they are non-zero and different, the answer is 6.

If the digits can be used more than once, then 900. If not, then 648.

Related questions

10,000

36

5040 different 4 digit numbers can be formed with the digits 123456789. This is assuming that no digits are repeated with each combination.

Numbers formed should be of at least 3 digits means they may be of 3 digits, 4 digits, 5 digits or 6 digits. There are 6 choices for digit in the units place. There are 5 and 4 choices for digits in ten and hundred’s place respectively. So, total number of ways by which 3 digits numbers can be formed = 6.5.4 = 120 Similarly, the total no.of ways by which 4 digits numbers can be formed = 6.5.4.3 = 360. the total no. of ways by which 5 digits numbers can be formed = 6.5.4.3.2 = 720. The total no.of ways by which 4 digits numbers can be formed = 6.5.4.3.2.1 = 720. So, total no.of ways by which the numbers of at least 3 digits can be formed = 120 + 360 + 720 + 720 = 1920.

If the digits are all different then 18. Otherwise, 192.

12 numbers.

100000 - including numbers with leading 0s

99999

Assuming you mean permutations of three digits, then the set of numbers that can be made with these digits is: 345 354 435 453 534 543 There are six possible permutations of three numbers.

how make the answer im don/now

Assuming they are non-zero and different, the answer is 6.

192, including ones containing repeat digits.

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