Advanced Math/lottery pick 3 numbers
QUESTION: How many sets of numbers are there from 1-999, that do not repeat in a sequence, like 123, repeating as 321 or 213, and do not have 2 of the same digit, like, 112, 242, 557
ANSWER: I'm pretty sure this is given by the usual formula for combinations. The fact that no 2 digits out of the 3 can be the same (2nd condition in your question) means that all 3 numbers are different so that we need to calculate how 3 different objects (numbers in this case) can be chosen out of 999. If sequences with the same numbers, but in a different order, are allowed, then we are looking for permutations, P, which is written (m out of n objects)
P(n,m) = n!/(n-m)!.
But since we don't want to count repeating sequences, we need to divide out the number of ways the m numbers can be arranged, or m!. These are called combinations and are written
C(n,m) = n1/[(n-m)!m!]
For 3 out of 999, we have
C(999,3) = 999!/[996!3!] = 999･998･997/(3･2) = 333･499･997 = 165,668,499.
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QUESTION: sorry I did not ask the question correctly. how many 3 digit combinations are there from 1-999, that have all 3 numbers different, and no 3 digit combination in any order is the same. thank you
ANSWER: Even with your re-phrasing of the question, I think my previous answer still holds. Maybe an example will help clarify how we are thinking of the problem.
Let's ask "how many 2-digit combinations are there from 1-5 that have both numbers different and none of the pairs are the same in any (either) order". I contend that the answer is
C(5,2) = 5!/[ (5-2)!2! ] = 10.
Enumerating the possibilities, we get
1 2 3 4 5 <- "objects" in our universal set; all are different
1 1 2
2 1 3
3 1 4
4 1 5
5 2 3
6 2 4
7 2 5
8 3 4
9 3 5
10 4 5.
Each row represents a unique pairing of the available objects and reversals are not double counted.
Is this not what you meant? Let me know. I'm confused.
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QUESTION: I want to pick lottery numbers for the pick 3 drawing. there are only 1000 combinations. I want to pick the least amount of numbers to have a better chance. 3 digit numbers with 2 digits the same, don't hit as often. I want to eliminate them. I am playing to win with the numbers boxed, which means they can come up in any order. So if I play the number 256, I don,at want to play 652,265,625,562, 526 because 256 would win, so eliminate I call them duplicate combinations. so how many combinations are left after removing the 3 digit combinations with duplicate numbers and the 3 digit combinations with duplicate combinations.
I think I get it now (I've never played the lottery, any lottery). You have 10 digits, 0-9, that you can choose to form a 3 digit set. So, without any restrictions, the number of possible sets is 1000, given by
(10 possible digits for the 1st pick)･(10 for the 2nd)･(10 for the 3rd) = 10･10･10 = 1000.
You want to avoid picking 2 numbers that are the same, which, for this 3-digit case, means that all 3 numbers are different. So the number of sets complying with this restriction are
(10 possible digits for the 1st pick)･(9 for the 2nd pick)･(8 for the 3rd) = 10･9･8 = 720 = 10!/(10-3)! = number of permutations.
You also want to bet the "boxed" variation where the order of the numbers doesn't count, meaning you want to reduce the possible number of 3-digit sets by the number of ways any 3 numbers can be arrange in a set. This is given by
(3 ways to pick the 1st number)･(2 ways to pick the 2nd once the first is picked)･( only one way to pick the last number since only 1 is left)
= 3! = 6.
Dividing out these extra combinations gives
720/6 = 120 = 10!/[ 7!3! ] = C(10,3).
This is the standard formula for choosing combinations of 3 objects out of 10.
120 is your answer.