That exhausts the possibilities, so there are 120+300=420 even four digit numbers that can be formed using the digits 0,1,2,3,4,5,6.
How many 4 digit numbers can be formed with the 10 digits 0 1 2 3 9 if a repetitions are allowed B repetitions are not allowed C the last digit must be zero and repetitions are not allowed?
(a)repetitions are allowed, Two ways to get the answer: 1. There are 9999 integers starting with 1 and ending with 9999. But the 999 integers starting with 1 and ending with 999 have less than 4 digits, so the desired number is 9999-999 or 9000 ways.
How many ordered codes can be formed using 4 of the digits in 1 2 3 4 and 5 if the digits are not repeated?
These 4 digits selected can be arranged in 4! ways. Hence the answer is 120 such numbers can be formed.
How many four digit even numbers can be formed using the digits 2 3 1 5 7 9 repetition of digits is not allowed?
In response to the question, there are 5*4*3*2 = 120 distinct permutations of 4 digit even numbers possible to be formed from the digits 1,3,5,6,8,9.
What is the greatest 4 digit even number?
Answer: 9998 is the right answer.
How many four digit even numbers can be formed using the digits 6 1 3 5 7 9 repetition of digits is not allowed?
Originally Answered: How many 4 digit even numbers can be formed from the digits 1, 3, 5, 6, 8 and 9 if repetition of digits is not allowed? the answer is 120 four-digit integers.
How many numbers can be formed from the digits 1 2 3 9 if repetition of digits is not allowed?
But we cannot take any of the 4 digits given in the second place because repetition is not allowed here. So, we have 3 intakes for second place. Similarly, for the 3rd place, we have to choose any one digit of the remaining two digits and proceed with 1 intake for the 4th place. Hence, 64 numbers can be formed.
