How many 3 digit numbers can be formed using the digits 1 2 3 4 and 5 where repetition of the digits is not allowed?
Problem 2: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated? Solution: Answer: 108. Let 3-digit number be XYZ.
How many 3 digit numbers can be formed using the digits 1 2 3 4 if repetition of digits are allowed in a number?
How many 3-digit numbers can be formed from the digits 1,2,3,4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition of the digits is not allowed? Total possible numbers =5×4×3=60.
How many three digit numbers with distinct digits can be formed such that the product of the digits is the cube of a positive integer?
The combination of three digits, whose product can give a perfect cube are (1,2,4) (1,3,9) (4,6,9) (3,8,9) and (2,4,8) when all three digits are distinct. With each of these combinations we can have 3! three digit numbers with distinct digits. Therefore, the number of such 3-digit numbers is 5(3!)
How many 3 digit numbers can be formed using the digits 2 3 and 4 without repetition?
Hence the answer is 900 such numbers can be formed.
How many 3 digits number with distinct digits can be formed if all the digits are in descending order?
We can make 120 three digit distinct numbers.
How many distinct three digit numbers can be formed using all the digits of 1/2 and 3?
Originally Answered: How many 3-digit numbers can be formed using only1, 2, 3? This is because you have space for three digits that can be chosen from the set [1,2,3]. So you can make 27 three digit different numbers .
How many 3 digit numbers can be formed in all ways?
Number of 3 digit number which can be formed = 9 × 8 × 7 = 504 ways
How many numbers can be formed from the second blank?
The second blank can now be filled by four remaining digits because repetition is not allowed and hence the digit selected for the first blank cannot be selected. So four ways. Similarly 3 for the third blank and two for the last blank. Hence the answer is 120 such number can be formed.
How many ways can you pick the first two and third digits?
So there are 6×6 or 36 ways to pick the first two digits. For each of the 6×6 or 36 ways we can pick the first two digits, we can pick the third digit any of the same 6 ways. So there are 6×6×6 or 216 ways to pick the three digits.
How do you solve as often as desired numbers?
I solved it this way because if a number needs to be formed with a certain number of digits (with no restrictions – which I think ‘as often as desired’ means), you assign a scale to the given constituent digits and then pick the same number of digits as the new number needs, going from highest to lowest.