How many 3 digit numbers can be formed using digit 0 1 2 3 4 5 which are even repetition not allowed?
Without repetition counting the number of 3digit numbers using digits from {0,1,2,3,4,5}: Count the number of 3-digit strings whose last digit is even. This gives 60−8=52 3-digit even numbers using digits from {0,1,2,3,4,5} without repetition.
How many 3 digit numbers can be created without repeating?
For each of these ways, y can be chosen in 8 ways. Hence, the desired number is 41 × 8 = 328 numbers 3-digit even numbers exist with no repetitions.
How do you find the 3rd digit greater than 330?
Finally For the third one greater than 330 we have two ways to achieve this either draw a 4, 5 or 6 for the first digit then we don’t care about the others making 3 × 6 × 5 = 90 Alternatively we must draw a 3 for our first digit then 4, 5 or 6 for our second and finally any digit making 1 × 3 × 5 = 15
How many ways can you fill a 3 digit number with 0?
A three-digit number cannot have 0 in hundreds (leftmost) place. So the hundreds place can be filled by any of the 6 digits (1,2,3,4,5 & 6) in 6 ways. Similarly, the tens place can be filled by any of the 6 digits ( including 0 but excluding the digit occupying the hundreds place) in 6 ways.
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 three digit numbers are there in 6/6/5?
So 6 × 6 × 5 = 180. Any of the 6 nonzero digits can be chosen for the hundreds position, and of the remaining 6 digits for the tens position, leaving 5 digits for the units position. So, there are (6) (5) (5) = 150 three digit numbers.