composing identification card with not repetition I saw a question in my textbook , it is trivial question and it has also a solution but i tried to solve it using completing rule.However , i could not reach the answer.The question is:
An identification card consists of $3$ digits selected from $10$ digits.
Find the probability that a randomly selected card contains the digits $1, 2,$
and $3$. Repetitions are $NOT$ permitted.
The solution is:
The number of permutations of $1, 2,$ and $3$ is $P(3,3)=6$
The number of permutations of three numbers in the set of $0, 1,..., 9$ is $P(10,3)=720$
Then , the solution is $\frac{6}{720}=0.008$
Everthing is cool, the solution is elegant.However , when i applied the complement rule, it did not work such that:
The number of identification cards which does not contain $1,2,3$ : $P(7,3)=210$
Then , the solution should be equal to $1- \frac{P(7,3)}{P(10,3)}=\frac{510}{720}=0.708$ ,but it is not..
What am i missing? Thanks for your helps..
 A: While complementing You eradicated $1,2,3$ from your set and Were Left with only digits $0,4,5,6,7,8,9$ which are in Total 7 . so You are making combinations Out of All of them So while doing this
You have forgot to Subtract Cases where one of the $(1,2,3)$ are in ID card numbers with some of these digits $0,4,5,6,7,8,9$ , For eg=> $(1,5,2)$ or maybe $(2,7,3)$, You still need to subtract these cases out for getting the desired results using Complementing :).
I think you can take it forward from this step.
A: Other answer (by @KartikBhatia, +1) and Comment (by @player3236) are correct that the complementary event must
allow one or two of the numbers $1,2,3.$ (Checking combinatorial results with alternative approaches
often catches mistakes; don't be discouraged from
doing that.)
Simulation: Additional confirmation of your first (correct) computation.
For ease of counting, designate unwanted 'digits' $4,5,6,7,8,9,0$
as (40, 50, ..., 100). Then the total of wanted digits $(1,2,3)$
must always be $6.$
A million iterations should give a probability
correct to three digits. The correct answer is $7/720 = 0.00833,$ simulated as $0.008263 \pm 0.000181.$
set.seed(2020)
nrs = c(c(1,2,3), seq(40,100,by=10))
id.123 = replicate(10^6, sum(sample(nrs,3))==6)
mean(id.123)
[1] 0.008263        # aprx 6/720
2*sd(id.123)/1000
[1] 0.0001810495    # 95% margin of sim error
6/720
[1] 0.008333333     # exact 6/720

