To quote the whole question accurately:

Various 3 digit numbers can be formed by permitting the digits 122. All arrangements are equally likely. Given that any resulting number is even, in how many ways can we arrange these digits to satisfy this condition?

The way I approached this problem is that I first fixed the last digit to be a 2 then got the permutations of the remaining 2 digits resulting in 2 ways: $$\{122, 212\}$$ However, my professor informed us that we should also multiply it by 2 since there are two '2's and we can interchange them giving us a total of 4 ways: $$\{122, 122, 212, 212\}$$ Where 122 and 122 are supposedly different since the 2's are interchanged. So my actual question is why should we distinguish between permutations that result in essentially the same number?

  • $\begingroup$ It depends on how you define your problem, are you looking for permutations or combinations. A permutation is an act of arranging the objects or numbers in order. Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter. $\endgroup$ Jul 11, 2022 at 9:02

1 Answer 1


To better understand the problem, you can think of having 3 balls {a, b, c} on which the numbers 1, 2 are written as {a -> 1, b -> 2, c -> 2}. So now you should think of the number of ways you can arrange these three balls in a way that the numbers written on them make an even number. Then the answer is: abc, acb, bac, cab equivalent to the {122,122,212,212} set you mentioned.

The reason you should distinguish between the permutations that result in the same number is that you will be interested in calculating the probability of a certain number happening (let's say by randomly drawing the balls from a bag) and some numbers are more probable simply because there are more ways you can create them.

  • $\begingroup$ So just to clarify: the only reason we would want unique arrangements instead of unique numbers is when we want to calculate the probability of forming a number by selecting the digits randomly from a bag. However, for the case that we only want the number of ways that it could be arranged to form an even number, it would only be 2 ways? $\endgroup$ Jul 11, 2022 at 10:19
  • $\begingroup$ I gave the probability example just to tell you why you might be interested in distinguishing between them. But the general answer would be that there could be multiple ways to make a unique number using a set of digits. Regardless of the goal, if someone asks you in how many ways you can make 122 from the set {1,2,2} you should be aware that 122 can be made in two different ways (remember the "abc" and "acb" example). $\endgroup$
    – Amin Shn
    Jul 11, 2022 at 11:38

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