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sondenso
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I think you forget to subtract double counting. We can divide your 36 ways to get a number by two groups.

1) First group, a pair group contains two '4' together like xx44xxx. We have 6 places and 6 ways to get such kind of a number. Moreover, this group doesn't contain double counting.

2) Second group, a separated group contains two '4' separated by other digit(s) like x4xx4xx. We have 36-6=30 ways to get such kind of a number. But in your reasoning you get the same number twice: x(first placed '4')xx(second placed '4')xx and x(second placed '4')xx(first placed '4')xx represent two ways but only one unique number.

Therefore, we should count numbers as 6+30/2=6+15=21
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Got it.
Thanks.

Could you explain about the 7 places at your first method?
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"Larry tried to type his new 7-digit phone number on a form ...." - 7 digits - 7 places for two '4'. Other 5 places are filled by 39269
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How do you solve such questions under pressure during the test?! :)
It took me more than 2 min to understand what the question was trying to say.

So: there are 6 ways to insert the first 4 and 7 ways to insert the second one. You need to divide the result by 2 as half of the time you will have the same number you already had. (6x7)/2= 21
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walker
A

1. 39269 contains 5 digits. Therefore, there are 2 '4' digits.
2. 2 '4' digits we can place in 7 possible positions by 7C2 different positions (we use 'C' instead of 'P' as a number like xx4x4x is insensitive to permutation of '4's)

7C2=7*6/2=21

Many thanks, OA is A



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