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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!

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    @ArturoMagidin thanks :)2012-04-16

2 Answers 2

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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 $\frac{80}{120}=\frac{2}{3}$.

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    sure, done!! :)2012-04-20
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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