0
Add your answer:
Earn +20 pts
How many 3 digit numbers can be formed using the digits 2 3 5 6 7 9 if repetition of digits is not allowed?
120 There are 6 digits in total. The numbers with 3 digits, with all digits distinct from each other, are the permutations of the 6 digits taken 3 at a time, and therefore there are 6*5*4 = 120 of them.
How many 6 digit numbers can be formed with 4 different digits?
The question can be re-stated as asking for the total number of permutations that can be derived from the following two groups of digits: AAABCD and AABBCD, where A, B, C and D are different. The number of ways of choosing the digit A, to be used three times, out of the ten digits {0, 1, 2, ... 9} is 10. Having done that, the number of ways of selecting 3 from the remaining 9 digits is 9C3 = (9*8*7)/(3*2*1) = 84. Thus there are 10*84 = 840 combinations of the form AAABCD. You now need all the distinct permutations of these 6 digits. The total number of permutations of 6 digits is 6! but because 3 of the digits are the same, these permutations are not all distinct. In fact, there are 6!/3! = 120 distinct permutations. That makes a total of 840*120 = 100800 such numbers. Next, the number of ways of choosing the digits A and B, each to be used twice, out of the ten digits {0, 1, 2, ... 9} is 10C2 = (10*9)/(2*1) = 45. Having done that, the number of ways of selecting C and D from the remaining 8 digits is 8C2 = (8*7)/(2*1) = 28. Thus there are 45*28 = 1260 combinations of the form AAABCD. You now need all the distinct permutations of these 6 digits. The total number of permutations of 6 digits is 6! but because 2 pairs of these digits are the same, these permutations are not all distinct. In fact, there are 6!/(2!*2!) = 180 distinct permutations. That makes a total of 1260*180 = 226800 such numbers. The grand total is, therefore, 100800 + 226800 = 327600 6-digit numbers made from 4 distinct digits.
How many three digits are there in which all the digit are distinct?
There are the digits 1 through 9 for the first digit. Then, we have 0 through 9 for the second digit - excluding the first digit. For the third digit, we have 0 through 9 excluding the two previous digits
How many three digit numbers with distinct digits?
1000there aren't even 1000 three-digitnumbers...There are the digits 1 through 9 for the first digit. Then, we have 0 through 9 for the second digit - excluding the first digit. For the third digit, we have 0 through 9, excluding the two previous digits for a total of 9*9*8 =81*8 =648 three digit numbers with distinct digits.EDIT- ... to be a little more specific, if your talking about a 3-digit password or unlock code its would be 1000. _-_
How many odd numbers with 4 distinct digits can be formed using the digits in 234567?
The last digit must be odd: three possibilities. The other digits can be anything else, as long as they're different: five times four times three possibilities. Thus, the answer is 3 x 5 x 4 x 3 = 180.
How many odd integers between 1000 and 9999 have distinct digits?
There are 120 of them.
What are 8 digit odd numbers with all digits different?
There are only five distinct odd digits.
What is a 4 digit ODD numbers have all 4 digits distinct?
5555
How many counting numbers have four distinct nonzero digits such that the sum of the four digits is 10?
1,2,3,4 1+2+3+4=10 4 times 3 times 2 times 1 =24 24 counting numbers
How many integers between 1000-9999 have distinct digits?
9*9*8*7 = 4536
How many positive integers less than 1000 have distinct digits?
10*9*8=720
How many 3 digit numbers can be formed using the digits 2 3 5 6 7 9 if repetition of digits is not allowed?
120 There are 6 digits in total. The numbers with 3 digits, with all digits distinct from each other, are the permutations of the 6 digits taken 3 at a time, and therefore there are 6*5*4 = 120 of them.
How many 6 digit numbers can be formed with 4 different digits?
The question can be re-stated as asking for the total number of permutations that can be derived from the following two groups of digits: AAABCD and AABBCD, where A, B, C and D are different. The number of ways of choosing the digit A, to be used three times, out of the ten digits {0, 1, 2, ... 9} is 10. Having done that, the number of ways of selecting 3 from the remaining 9 digits is 9C3 = (9*8*7)/(3*2*1) = 84. Thus there are 10*84 = 840 combinations of the form AAABCD. You now need all the distinct permutations of these 6 digits. The total number of permutations of 6 digits is 6! but because 3 of the digits are the same, these permutations are not all distinct. In fact, there are 6!/3! = 120 distinct permutations. That makes a total of 840*120 = 100800 such numbers. Next, the number of ways of choosing the digits A and B, each to be used twice, out of the ten digits {0, 1, 2, ... 9} is 10C2 = (10*9)/(2*1) = 45. Having done that, the number of ways of selecting C and D from the remaining 8 digits is 8C2 = (8*7)/(2*1) = 28. Thus there are 45*28 = 1260 combinations of the form AAABCD. You now need all the distinct permutations of these 6 digits. The total number of permutations of 6 digits is 6! but because 2 pairs of these digits are the same, these permutations are not all distinct. In fact, there are 6!/(2!*2!) = 180 distinct permutations. That makes a total of 1260*180 = 226800 such numbers. The grand total is, therefore, 100800 + 226800 = 327600 6-digit numbers made from 4 distinct digits.
How many license plates with 3 decimal digits followed by 3 letters can you build if you are not allowed to repeat digits and you are not allowed to repeat letters?
3 decimal digits without repeats can form (10 x 9 x 8) = 720 distinct displays.For each of these . . .3 letters without repeats can form (26 x 25 x 24) = 15,600 distinct displays.Combine them on one plate, and there are (720) x (15,600) = 11,232,000 distinct displays available.
What is the smallest 6-digit number using three distinct digits?
100122 or, if negative numbers are permitted, -998877
How many counting numbers have four distinct nonzero digits that have the sum of the four digits is 11?
1235; 1253; 1325; 1352; 1523; 1532; 2135; 2153; 2315; 2351; 2513; 2531; 3125; 3152; 3215; 3251; 3512; 3521; 5123; 5132; 5213; 5231; 5312; 5321. These are 24 such numbers which have four distinct nonzero digits and add up to 11.
Find the all distinct four-digit numbers that contain only the digits 12345?
There are 625 of them - too many to list.
