How many 3 digit numbers can be formed by using the digits 0 1 3 5 7 when the digits may be repeated any number of times?

100 3 digit numbers
Therefore, a total of 100 3 digit numbers can be formed using the digits 0, 1, 3, 5, 7 when repetition is allowed.

How many 3 digit numbers are there which contain digit 5?

Total three digit numbers having one or more 5 are 72+72+81=225 numbers.

How many 3 digit numbers can be formed by using the digits?

Hence 1, 7, 8, 9. The value of r will be 3, as we need a form 3 digit number only. Hence, 24 3-digits numbers can be formed without using the digits 0, 2, 3, 4, 5 and 6.

How many 3 digit numbers are there when a digit may be repeated any numbers of time?

How many 3-digit numbers are there when a digit may be repeated any number of times? We may fill hundred’s digit by any digit from 1 to 9 , i.e., in 9 ways. Each of the ten’s and unit’s digits may be filled in 10 ways. Required number of numbers =(9×10×10)=900.

How many 3 digit numbers can be formed by using the digits 0 5 7 without repeating any digit in the number?

The possible 3-digit numbers are: 507, 705, 750, 570.

How many 3 digit numbers have at least one 3 as a digit?

252
So there are 8 x 9 x 9 of these. That’s 648 altogether. So the number of 3-digit numbers with at least one three is 900 – 648 = 252.

How many 3 digit numbers can be formed using the digits 0?

Originally Answered: How many 3 digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if repetitions of digits are not allowed? There are nine possible first digits, because numbers beginning in 0 drop the 0, so 012 is really just 12, a two digit number. The first digit must be 1 thru 9, nine possible digits.

What is the total number of 3 digit odd numbers?

So for 1 in the units place, the number of 3-digited numbers is 5*5=25. The same analogy holds for 3 & 5 occupying the units place. Thus the total number of 3 digited odd numbers are 25+25+25=75. Count of 3-digit numbers with digits HTU all different formed from 7 digits 0 1 2 3 4 5 6 is 6×6×5=180 as digit choices for H T U are 6 6 5 respectively.

How many ways can you choose the second and third digit?

Choose the second digit any of 3 ways, 4,5, or 6 That leaves 5 unchosen digits Choose the third digit any of these 5 ways. Answer case 1: (1) (3) (5) = 15 ways.

How many unchosen digits if each digit can be used only once?

0,1,2,3,4,5,6 if each digit can be used only once? There are 7 unchosen digits Choose the first digit any of 6 ways other than 0. That leaves 6 unchosen digits Choose the second digit any of 6 ways.