Explanation

Now here we have digits 3,4 and 8. How many ways can we arrange them to form a three digit number.

Lets pick one digit at a time.

1st Digit -> 3 Possibilities
2nd Digit -> 2 Possibilities
3rd Digit -> 1 Possibility

Number of digits = \(3 \times 2 \times 1 =6.\)

Hence option B is correct


The other way to solve it is to write out the possible three-digit positive integers and count them: 348, 384, 438, 483, 834, 843 → 6 three-digit positive integers While this method is efficient when the question only has three digits, it can get quite complicated with more digits or four-digit positive integers.


Note this is applicable for the statement:
How many ways can n distinct things be arranged? Answer: \(n!\)
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However this could also be possible in this question that the repetition of digit is aloud which is not clearly mentioned in the question. One can find for both the cases and then you will come to know that your answer is correct for which cases by scrutinizing the choices properly.

Normally you would have a point there, however, the question states 'three digits are' implying that all three digits exist in the number and since there are only 3 places there is no scope for repetition.

If the question was formatted as 'total number of 3 digit numbers formed by using 3, 4 and 8' then repetition is a definite possibility.

Paying attention to the language of the questions in the GRE is extremely important.

Hope this helps.
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3! = 6

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