I've started studying permutations and I'm confused as to how to properly solve a problem. For example, select 4 digits from the set $\{1,2,3,4,5,6,7\}$ without replacement. What is the probability that the number formed is even?

Looking at this question it's pretty obvious that the answer is $\frac 37$ since there are 3 even numbers from the set so the last digit (ones digit) can be any of the even numbers. However, if I approach this problem step by step I cannot get the same answer:

To solve this problem step by step, let $A$ be the event that the number is even and S represent the sample space. Let $n(s)$ be the number of ways any number can be formed and $n(A)$ be the number of ways an even number can be formed.

$n(s) = 7^{(4)}$ since the first digit (the thousands digit) can be chosen from 7 the next from 6... Now I'm confused on how to calculate $n(A)$ since the first (thousands), second(hundreds), or third digit(tens) can be an even number. I calculated $n(A)$ as following: $n(A) = (7*6*5*3) + (7*6*5*2) + (7*6*5*1)$ the first group is when the first three digits are odd, the second group is for when one of the digits is odd and the last group is when two digits are odd. But my dilemma is that now $n(A) > n(s)$ which is impossible so I did something wrong.


1 Answer 1


Your notation is a bit strange. For instance, what is $7^{(4)}$? Also, why do you consider it necessary to count all the permutations of the other digits? Just consider the partition of the sample space where we only look at the first digit since the number is even if and only if the first digit is even.

In any case, the permutations that correspond to the first digit being even are $3 \cdot 6 \cdot 5 \cdot 4$ in number (the first must be even, the next can be any other value, etc.), and the total number of permutations is $7 \cdot 6 \cdot 5 \cdot 4$. If you divide the former by the latter you get $3/7$.

  • $\begingroup$ This is a solution for the first sampled number being even. I thought the question was asking "what are the odds of the fourth number being even?" $\endgroup$
    – Adam C
    Commented Aug 10, 2015 at 18:02
  • $\begingroup$ "What is the probability that the number formed is even?" $\endgroup$
    – dsaxton
    Commented Aug 10, 2015 at 18:03
  • $\begingroup$ If you pull {1,2,3,4} you can either form 1234 or 4321. That's why the questioner was concerned with "the first, second, or third digit can be even." $\endgroup$
    – Adam C
    Commented Aug 10, 2015 at 18:06
  • $\begingroup$ I think it's understood that we're sampling the digits in order, otherwise the question makes no sense. Is $\{1, 2, 3, 4 \}$ even or odd? This has no answer if the selection can correspond to either $1234$ or $4321$. Also what would be the relevance of the first, second or third digit being even? $\endgroup$
    – dsaxton
    Commented Aug 10, 2015 at 18:12
  • $\begingroup$ dsaxton I am a bit confused as well with the statement "since the number is even if and only if the first digit is even." A number where the first digit is even does not mean the number itself will be even for example $4321$ is an odd number with an even first digit. When I wrote $7^{(4)}$ I meant $7 \cdot 6 \cdot 5 \cdot 4$. $\endgroup$
    – Daniel
    Commented Aug 10, 2015 at 18:13

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