0
Add your answer:
Earn +20 pts
How many permutations of 3 different digits are there from the ten digits 0 to 9 inc?
The answer is 10P3 = 10!/(10-3)! = 10*9*8 = 720
How many four digit numbers can be formed from digits 1 2 3 4 if repeated is allowed and how many are odd numbers?
This is permutations with repetition. The answer is 4^4 = 256 total permutations. Since 2 of the digits used are odd (and 2 are even), then half of the possibilities will be odd: 128 odd numbers.
How many permutations can you get with 3 letters and 3 digits in the following format AAA000 - ZZZ999?
The answer is 26*26*26*10*10*10 = 17,576,000
How many license plates can be made using three digits and three letters without repetition and alternating letters and digits?
If it alternates and you start with a letter, then there are 11,232,000 permutations. Then if you start with a number and alternate, there are another 11,232,000 permutations, for a total of 22,464,000 permutations. If you exclude the letters I, S, B, and O (because they look kind-of like 1, 5, 8, & 0 - kind-of important on license plates) then you are down to 6,652,800 & 13,305,600 respectively.
How do you unlock 5 digits lock?
If you wish to use a brute-force method, assuming the digits go from 0 to 9 there are 10,000 possible permutations.
How many permutations of 3 different digits are there chosen from the ten digits 0 to 9 inclusive?
How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?
How many different numbers can be formed with the digits 345?
Assuming you mean permutations of three digits, then the set of numbers that can be made with these digits is: 345 354 435 453 534 543 There are six possible permutations of three numbers.
How many permutations of 3 different digits are there from the ten digits 0 to 9 inc?
The answer is 10P3 = 10!/(10-3)! = 10*9*8 = 720
How many numbers used for 10000 permutation?
8 digits will generate over 40,000 permutations.
How many four digit numbers can be formed from digits 1 2 3 4 if repeated is allowed and how many are odd numbers?
This is permutations with repetition. The answer is 4^4 = 256 total permutations. Since 2 of the digits used are odd (and 2 are even), then half of the possibilities will be odd: 128 odd numbers.
How many permutations of two different digits can be obtained from a set of four different digits?
6 of them. 4C2 = 4!/(2!*2!) = 4*3/(2*1) = 6
How many possible 5 digit combInations are there with in ten digits?
In case of digits or numbers, when we say combination, since the order of the numbers matters, we really need to find the permutations. For example, 12 and 21 are two permutations and not combination. The permutation for 10 digits chosen 5 ways is 10*9*8*7*6, which is 30240. But how many are there in whole like four digit has 24.
How many number can be created with 15679?
It is possible to create infinitely many numbers, of infinitely many different lengths, using the digits of the given number. Using each of the digits, and only once, there are 5! = 120 different permutations.
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 permutations can you get with 3 letters and 3 digits in the following format AAA000 - ZZZ999?
The answer is 26*26*26*10*10*10 = 17,576,000
How many 4 digit combinations can be made from 6 digits?
Oh, dude, you're hitting me with some math vibes here. So, if you have 6 digits to choose from to make a 4-digit combination, you can calculate that by using the formula for permutations: 6P4, which equals 360. So, like, you can make 360 different 4-digit combinations from those 6 digits. Math is wild, man.
How many 4 number combinations using digits 1357 one and only once?
There is only one combination. The order of the digits in combinations makes no difference. They are considered as being different if they are permutations, not combinations.
