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The digits from 1 to 6 are arranged to form a six-digit multiple of 5. What is the probability that the number is greater than 300,000? Express your answer as a common fraction.

 Dec 15, 2020
 #1
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"The digits from 1 to 6 are arranged to form a six-digit multiple of 5. What is the probability that the number is greater than 300,000? Express your answer as a common fraction."

 

The number must end in 5 if it's to be a multiple of 5.

That leaves 5 possibilites for the first digit, for each of which there are 4 possibilities for the second digit, for each of which ...etc., making the total number of possible numbers N = 5!

 

Any number beginning with 3, 4 or 6 will be bigger than 300,000.  

With 3 as the first number there are 4! possible ways of distrbuting the others.

Similary, with 4 as the first there are 4! ways, and with 6 as the first ther are 4! ways.

Hence there are n = 3*4! possible numbers larger than 300,000.

 

The probability is therefore p = n/N  or  p = 3*4!/5!  or  p = 3/5.

 Dec 15, 2020
 #2
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Since there are 6 numbers that must be arranged in such a way that each arrangement must end in 5 in order to be a multiple of 5. Then, you would have: [6 - 1]! =5! = 120 permutations. I'm assuming that repeats are NOT allowed. In order for an arrangement to be greater than 300,000, it must necessarily begin with either: 3, 4, or 6 and end in 5. So, the number of arrangements that would be greater than 300,000 are: 3 x 4! =72.

 

Therefore, the probability is: 72 / 120 == 3 / 5

 Dec 15, 2020

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