Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was doing this question:

Repetitions not allowed: Using the following six digits: 2, 3, 5, 6, 7, 9 What is the probability that a three-digit number greater than 400 will be formed from these six digits?

I understand to get the numerator you have to do the following:

P(6,3) 3-digit numbers $6!/(6-3)!= 6\times 5\times 4 = 120$

And to get the denominator you have to do this

number of ways form a number greater than 400

But after that I’m completely lost, I understand there a 4 choices but that’s it. Could someone help me on the methodology of using these four choices?

Thanks in advance!

share|cite|improve this question
For numbers greater than $400$, there are four possibilities for the first digit, five left for the second digit, and four left for the third digit. Total number, $4\times 5\times 4$. – Arturo Magidin Apr 16 '12 at 16:38
You have numerator and denominator backwards. You will get a probability greater than 1 this way. – Ross Millikan Apr 16 '12 at 16:41
@ArturoMagidin thanks but how did you see that there four, five and four possibilities? – Xabi Apr 16 '12 at 16:41
@RossMillikan my mistake but i meant the other way round – Xabi Apr 16 '12 at 16:43
@Fatz: There are four possibilities for the first digit to ensure you get more than 400: first digit must be either 5, 6, 7, or 9. Once you use up that number, it doesn't matter what the other digits are, you will get a number greater than 400. So there are 5 possibilities for the second digit (any of the original six except for the one used in the first digit), and four for the third digit (any of the original six, except for the two that we used in the first and second digit). – Arturo Magidin Apr 16 '12 at 16:43
up vote 2 down vote accepted

I'm assuming each digit chosen from $\{ 2,3,5,6,7,9\}$ must be distinct. For the first digit, you must choose a number $\ge$ 4, so you have four choices. After that, any number will do, as it only needs to be $\ge 0$. So you have $6-1=5$ choices, and similarly for the last digit you have $5-1=4$ choices, for a total of $4 \times 5 \times 4=80$ choices. The total number of choices is $6 \times 5 \times 4 = \frac{6!}{3!}=120$. Therefore the probability of choosing a number greater than 400 is $\frac{80}{120}=\frac{2}{3}$.

share|cite|improve this answer
Thanks! @Jackson Walters – Xabi Apr 16 '12 at 16:48
No problem, please accept if this is what you were looking for. – Jackson Apr 16 '12 at 17:42
sure, done!! :) – Xabi Apr 20 '12 at 16:25

The simple solution is to note that only the first digit matters. If it is 5,6,7, or 9, the resulting number will be greater than 400. So 4/6=2/3

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.