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# The digits 3, 4, 5, 6 and 7 are randomly arranged to form a three-digit number. Digits are not repeated. What is the probability that the number is even and greater than 700?

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Dec 14, 2017

12/120=1/10=10%

#### Explanation:

There are 5! = 120 ways to randomly arrange the 5 digits.

To satisfy the conditions, we want the first digit to be 7. We want the last digit to either be 4 or 6. With the 7 forced to be the first digit, we have 4 digits to account for.

To get an even digit, we have a population of 2 and we are choosing 1:

"^nC_k=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$

"^2C_1=(2!)/((1!)(2-1)!)=2! =2

We can now account for the remaining 3 digits, which is 3! = 6.

Therefore, the number of ways to get an even number greater than 700 is:

$2 \times 6 = 12$

and the probability of randomly arranging numbers into this arrangement is:

12/120=1/10=10%

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