Hi Mike. If order mattered, we would say select the first number (10 choices), then the second (9 choices) then the third (8 choices). So there would be 10 x 9 x 8 = 720 possible choices. But because order does not matter, we have to take care of duplicates as you mentioned. How many duplicates are there for each set of three numbers? Well, again, we can choose one of the three as the "first", so there are 3 choices for that, then 2 choices for the "second" digit, then 1 choice for the last digit. There are, you see, 3 x 2 x 1 = 6 possible ways of arranging the three digits. Therefore in that set of 720 possibilities, each unique combination of three digits is represented 6 times. So we just divide by 6. 720 / 6 = 120. That's your answer. These are what are mathematically called "combinations". You can use a formula involving factorials to determine the number of combinations. In this case, we say this is "10 Choose 3" and write it as 10C3. That means from a set of ten (digits in
this case), choose 3 regardless of order. The formula for nCm is nCm = n! / [ m! x (n-m)!] So in your question, we have 10C3 = 10! / [3! x (10-3)!] = 10! / [3! 7!]
= (10 x 9 x 8 x 7!) / [ (3 x 2 x 1) 7!]
= (720) / 6 ... because the 7! on the top and bottom cancels.
= 120 as we expected. For more information on Combinations and their close cousins Permutations (where order matters), look up these words in our Quick Search. Cheers,
Steve La Rocque. In December 2021 we received a note from Doug to say that he thought Steve's answer to Mike was incorrect. He felt so since Steve didn't allow combinations such as 000, 111, or 011. In Mike's question he said "The numbers can be repeated as long as they are not of the same set." and Steve interpreted that to mean that the combinations Doug mentioned were not valid. If Steve's interpretation is correct then there are 120 valid combinations. If however such combinations are allowed then there are 10 combinations with the same digit repeated three times, 000, 111, ... , 999. For combinations with a digit repeated twice and a different digit in the remaining position there are 10 choices for the digit to be repeated twice and then 9 choices for the third digit giving 10x9=90 combinations. Hence, as Dough pointed out, in this case there are 120 + 10 + 90 = 220 valid combinations. | FAQs
We have to find the number of arrangements of 3 digits which can be formed from digits 0 to 9. Therefore, in 1000 arrangements 3 digits can be formed from the digits 0 through 9.
How many 3 digit numbers can be formed using 0 to 9? ›
The units place can be filled with 0 to 9 ways. i.e., 10 ways. So a total of 9 * 10* 10 = 900 three digit numbers can be formed.
How many 3 number combinations add up to 9? ›
We see that 9 can be written as the sum of 1, 2 and 6; 1, 3, and 5; or 2, 3 and 4. For each of these groups of three integers the addends can be arranged in 3! = 6 different ways. Therefore, the total number of ways that 9 can be written as a sum of three distinct positive integers is 6 × 3 = 18 ways.
What are the possibilities for a 3 digit code? ›
The possible codes for a 3-digit lock from 0 to 9 are all the combinations of three digits ranging from 000 to 999. There are a total of 1,000 possible combinations [1].
What is the most common 3 digit code? ›
Expert-Verified Answer
Answer: The most common 3 digit number is 100. Step 1: A 3 digit number is any number between 100 and 999. Step 2: The most common 3 digit number is 100 because it is the lowest possible 3 digit number, so it has the highest frequency of occurrence.
How many combinations are there from 0 to 9? ›
Number of combinations for a 4 digit code (0-9) are 10*10*10*10=10,000, which can easily be brute forced.
How many 3 digit even numbers can be formed using 0 9 without repetition? ›
If z is 0, then x has 9 choices. Therefore, z and x can be chosen in (1 × 9) + (4 × 8) = 41 ways. For each of these ways, y can be chosen in 8 ways. Hence, the desired number is 41 × 8 = 328 numbers 3-digit even numbers exist with no repetitions.
How many 3-digit numbers can be formed from the digits 1234567890? ›
The answer is 10C3 = 10*9*8/(3*2*1) = 120 combinations.
How many 3-digit numbers can be formed without repetition? ›
Hence, the total number of required numbers= 9×9×8=648.
How many 3 digit numbers are there in all? ›
List of Three Digits Numbers. As we already discussed, the three digits numbers are from 100 (hundred) to 999 (nine hundred ninety-nine). Before we see the list of three digits numbers, first let us discuss how many three digits numbers we have from 100 to 999. Hence, there are 900 three-digit numbers in total.
The formulas for permutations, combinations without repetition, and combinations with repetition are given by P ( n , r ) = n ! ( n − r ) ! , C ( n , r ) = P ( n , r ) r !
What are the chances of guessing a 3 digit code first try? ›
Expert-Verified Answer
The probability that the player guesses the correct code on the first try is 0.001 or 0.1%.
How many permutations of 3 different digits are there chosen from the digits 0 to 9 inclusive? ›
There are, you see, 3 x 2 x 1 = 6 possible ways of arranging the three digits. Therefore in that set of 720 possibilities, each unique combination of three digits is represented 6 times.
How many 3-digit even numbers can be found from 0 to 9 if repetition is not allowed? ›
There are 328 three-digit even numbers with distinct digits.
How many 3 digit numbers can be formed by using the digits 1 to 9 without repetition? ›
Hence, the required number of numbers =504.
How many 3-digit numbers can be formed using 1/9? ›
⇒So, the required number of ways in which three-digit numbers can be formed from the given digits is 9×8×7=504. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
