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How many 4 digit numbers can be formed using 0 1 2 3 5 7 how many contain 2 in units place?
Wiki User
∙ 12y ago
Without repetition, 300. Of these 48 will have a 2 in the units place.
Wiki User
∙ 12y ago
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What numbers is formed if you increase by 1 the digit in the given place value of 568905?
It depends on which place value is given!
How many four digit numbers can be formed using all digits if same digit on four place is not allowed example 5555 is not allowed?
Total possible 4-digit numbers= 1000, 1001,...,9999 = 9000 Total with same digit numbers = 1111,2222,...,9999 = 9 9000 - 9 = 8991
What is the number if there is a 4 digit number no numbers are repeated the digit is odd the digit in the tens place is three times the digit in the thousands place and the sum of the numbers is 27?
3897
How many 4 digit numbers can you make with 5 3 1 7?
256 Each "place holder" if you will, may contain 4 numbers. So it is 4x4x4x4
What is remainder if the face value of the digit at the lakhs place is subtracted from the place value of the digit at the lakhs place in the smallest seven-digit number formed by using all the digits?
199998
What numbers is formed if you increase by 1 the digit in the given place value of 568905?
It depends on which place value is given!
How many four digit numbers can be formed using all digits if same digit on four place is not allowed example 5555 is not allowed?
Total possible 4-digit numbers= 1000, 1001,...,9999 = 9000 Total with same digit numbers = 1111,2222,...,9999 = 9 9000 - 9 = 8991
What is the number if there is a 4 digit number no numbers are repeated the digit is odd the digit in the tens place is three times the digit in the thousands place and the sum of the numbers is 27?
3897
How many 4 digit numbers can you make with 5 3 1 7?
256 Each "place holder" if you will, may contain 4 numbers. So it is 4x4x4x4
How do you multiply by two-digit numbers?
To multiply two digit numbers, multiply each place value of a factor by each place value digit and add the results.
How many numbers greater than 4000 can be formed with the digits 2 3 4 5 and 6 if no digit is used more than once in a number?
The numbers between 4,000 and 10,000 _ _ _ _ 1st place only can filled by 4, 5, and 6 2nd place only can filled by the other 4 digit 3rd place only can filled by the other 3 digit 4th place only can filled by the other 2 digit So: 3 x 4 x 3 x 2 = 72 numbers The numbers greater than 10,000 5 x 4 x 3 x 2 x 1 = 120 numbers The total is 72 + 120 = 192 numbers can formed.
What is remainder if the face value of the digit at the lakhs place is subtracted from the place value of the digit at the lakhs place in the smallest seven-digit number formed by using all the digits?
199998
How many 3 digit numbers have at least one repeated digit?
There are a total of 900 3-digt numbers (100 thru 999); Of these there are 9x9x8 = 648 numbers that do not contain repeating digits (first position: 9 possibilities, second place 9 possibilities; the 10 digits minus the one used in the first place. and third place 8 possibilities; 10 digits minus the two used in the first and second places) So, there should be 900-648=252 3-digit numbers with at least one repeating digit.
What are the difference between place value of whole numbers and decimal numbers?
There is no difference. In both cases, the place value, of any digit, is ten times the place value of the digit to its right.
What are the greatest and least 6 digit numbers in which the digit in the ten thousands place is twice the digit in the tens place?
Greatest: 989949 Least: 100000
How many three digit numbers are there with zero at tens place?
there are no 3 digit tis two digit! :) * * * * * 90 of them.
How many numbers between 1 and 999999 contain at least one 1?
We are to calculate the number of 1-digit to 6-digit numbers (i.e. numbers less than one million) that have at least one 1. Let's divide this into six cases by the number of digits in the number.For one digit numbers, the only number is 1 - 1 such number.For two digit numbers, we can find the number which do not contain a 1 more easily. There are 8 possible digits for the tens place (0 and 1 are not allowed), and 9 digits for the ones place (1 is not allowed). Thus we have 8 * 9 = 72 which do not work. There are 90 two digit numbers, so 90 - 72 = 18 such numbers.For three digit numbers, we will proceed in the same manner. There are 900 total 3 digit numbers, of which 8 * 9 * 9 = 648 numbers which do not contain a 1. The remaining 900 - 648 = 252 numbers contain at least 1 one - 252.For four digit numbers, continuing in the same manner we have 8 * 9 * 9 * 9 = 5832 which are not good out of the 9000 total. So, 9000 - 5832 = 3168.For five digit numbers, we have 8 * 9 * 9 * 9 * 9 = 52488. The remaining 90000 - 52488 = 37512 are good.For six digit numbers, we have 8 * 9 * 9 * 9 * 9 * 9 = 472392. The reamining 900000 - 472392 = 427608 are good.Finally, adding all of the totals, we have 1 + 18 + 252 + 3168 + 37512 + 427608 = 468,559 numbers between 1 and 999999 which contain at least 1 one.
