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M26-19

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Re M26-19 [#permalink]

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New post 16 Jun 2017, 21:51
I think this is a high-quality question and I don't agree with the explanation. The answer should be 1530/10^5. The explanation does not consider the number of arrangements that are possible in the code. Please get this corrected.

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Re: M26-19 [#permalink]

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New post 17 Jun 2017, 03:24
mohitbagga2006@gmail.com wrote:
I think this is a high-quality question and I don't agree with the explanation. The answer should be 1530/10^5. The explanation does not consider the number of arrangements that are possible in the code. Please get this corrected.


My friend, the answer is 100% correct. You can check more here: https://gmatclub.com/forum/baker-s-dozen-128782.html or just google the question.

Hope it helps.
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Re M26-19 [#permalink]

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New post 29 Sep 2017, 22:31
to find # of ways to have exactly 3 digits 6 i used the formula 9*9*5P5/3! ( the order is important, 12 not equal 21, we also need to divide by 3!, because we have 3 indistinguishable numbers ( 3 digits of 6). The solutions is 9*9*5C3. Could you hep me to understand what my mistake is.

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Re: M26-19 [#permalink]

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New post 03 Nov 2017, 01:15
Guys,

After spending sometime on the question, I could understand why the answer is 810 but not 1530. Here is my explanation.

Number of ways in which 3 places are occupied by the digit 6 and the remaining 2 places are occupied by two different digits , say 1 and 2

Now, let us see what happens when we calculate 9*8*5!/(3!) = 1440 ways

-> 9*8 implies we are considering all the arrangements pertaining to two digits. As we took digits 1 and 2 as our examples, we happen to consider the arrangements 1 2 and 2 1 separately

Now let us consider the arrangement 1 2 followed by three 6's. when you multiply this arrangement of 1 2 6 6 6 with 5!/(3!), you consider the cases including the ones in which 2 comes before 1 . For example 2 1 6 6 6 is also considered as 5!/(3!) includes all the possible arrangements.

But as explained earlier, since 9*8 considers the arrangement 2 1 as distinct from 1 2 , you therefore multiply the arrangement 2 1 6 6 6 with 5!/(3!) separately, resulting in the double counting since all the arrangements pertaining to three 6's and 1,2 are already counted.

Therefore you should divide the computation 9*8*5!/(3!) by 2! inorder to avoid the double counting.

So the answer is 720+ 90 (Number of ways in which two other digits are same i.e 9*1*5!/(3!*2!)) = 810

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Re: M26-19 [#permalink]

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New post 21 Nov 2017, 23:59
Bunuel wrote:
Official Solution:

A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digits 6?

A. \(\frac{860}{90,000}\)
B. \(\frac{810}{100,000}\)
C. \(\frac{858}{100,000}\)
D. \(\frac{860}{100,000}\)
E. \(\frac{1530}{100,000}\)


Total # of 5-digit codes is \(10^5\), notice that it's not \(9*10^4\), since in a code we can have zero as the first digit.

# of passwords with three digits 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have \(9*9\) and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\).


Answer: B


Hi Bunuel,
I did exactly the same except I could not do the last part i.e arrangement of 3-6's among 5 places. Am I correct in saying that we have to find the total no of arrangements for A666A & A666B? Where A and B are different digits. So in this case {5!/(3!2!) + 5!/3!} depending on whether remaining two digits are same or different.

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Re: M26-19 [#permalink]

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New post 10 Dec 2017, 09:12
Hi Bunuel,
I did not understand below part.Could you please explain??How

# of passwords with three digits 6 is 9∗9∗C35=8109∗9∗C53=810: each out of two other digits (not 6) has 9 choices, thus we have 9∗99∗9 and C35C53 is ways to choose which 3 digits will be 6's out of 5 digits we have.

Thanks
Abhinas

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Re: M26-19   [#permalink] 10 Dec 2017, 09:12

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