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M26-19

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M26-19  [#permalink]

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New post 16 Sep 2014, 00:25
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Question Stats:

58% (02:22) correct 42% (02:27) wrong based on 163 sessions

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Re M26-19  [#permalink]

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New post 16 Sep 2014, 00:25
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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
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Re: M26-19  [#permalink]

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New post 05 Nov 2016, 22:00
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Hi guys,

Alternate solution that worked for me - let me know what you think!

1. The probability of choosing a 6 is (1/10)
2. There are 5 digits, so the combined probability is (1/10)*(1/10)*(1/10)*(9/10)*(9/10) = 81/100,000
3. There are 5C3 ways to organize the 3 6's you picked

The answer is 10*(81/100,000) = 810/100,000
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M26-19  [#permalink]

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New post 05 May 2016, 13:21
Pretty sure it should be 1/100 since the # of passwords with three digits 6 should be 10∗10∗C35=10*10*10=1000

Each out of two other digits (not 6) should have 10 choices if you are considering 10 possible digits when calculating the total possible outcomes for the denominator (10,000) in the first step.
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New post 10 May 2016, 19:55
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wmichaelxie wrote:
Pretty sure it should be 1/100 since the # of passwords with three digits 6 should be 10∗10∗C35=10*10*10=1000

Each out of two other digits (not 6) should have 10 choices if you are considering 10 possible digits when calculating the total possible outcomes for the denominator (10,000) in the first step.


Hi you are wrong here..
If you are considering other digits than 6... total digits = 10-1 = 9..
so possiblity = 9*9*5C3 = 810.. as explained by others..

the way you are taking, you can have a 5-digit code with all 6 - 66666 or 4 6s - 66661 or 16666 etc..

but Q asks us EXACTLy 3...
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New post 03 Jun 2016, 09:16
I understand that 5C3 is how many ways that the digit '6' is allocated in each of the five slots, but what about the other two digits? Where do you account for the number of ways that those two digits can be allocated in each of the five slots?
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New post 05 Jun 2016, 03:52
glochou wrote:
I understand that 5C3 is how many ways that the digit '6' is allocated in each of the five slots, but what about the other two digits? Where do you account for the number of ways that those two digits can be allocated in each of the five slots?


5C3 is the number of ways to choose which 3 digits will be 6's out of 5 digits we have. The remaining 2 places will be taken by non-sixes (9*9 combination).
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Re: M26-19  [#permalink]

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New post 03 Nov 2017, 01:15
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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 30 Jun 2019, 09:36
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Bunuel wrote:
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. \(\frac{1530}{100,000}\)


We're going to apply the MISSISSIPPI rule at the end of my solution, so here's what it says:
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]

-----NOW ONTO THE QUESTION------------------------------

We'll consider two cases:
case i: We have three 6's and two DIFFERENT digits (e.g., 66612)
case ii: We have three 6's and two IDENTICAL digits (e.g., 66677)

case i: We have three 6's and two DIFFERENT digits (e.g., 66612)
We already have three 6's. So, we must select 2 different digits from (0,1,2,3,4,5,7,8 and 9)
We can do this in 9C2 ways (=36 ways)
Now that we've selected our 5 digits, we must ARRANGE them, which means we can use the MISSISSIPPI rule.
We can arrange 3 identical digits and 2 different digits in 5!/3! ways = 20 ways
So, we can have three 6's and two DIFFERENT digits in (36)(20) ways (= 720 ways)


case ii: We have three 6's and two IDENTICAL digits (e.g., 66677)
We already have three 6's. So, we must select 1 digit (which we'll duplicate)
Since we're selecting 1 digit from (0,1,2,3,4,5,7,8,9) we can do so in 9 ways
Now that we've selected our 5 digits, we must ARRANGE them, which means we can use the MISSISSIPPI rule.
We can arrange 3 identical 6's and 2 other identical digits in 5!/3!2! ways = 10 ways
So, we can have three 6's and two IDENTICAL digits in (9)(10) ways (= 90 ways)

So, TOTAL number of ways to have three 6's = 720 + 90 = 810

Since there are 100,000 possible 5-digit codes, P(having exactly three 6's) = 810/100,000

Answer: B

Cheers,
Brent
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New post 30 Jul 2019, 09:28
Another (slightly more painful) option here if you're struggling with the repeated terms:

First, carefully work out all of the possible arrangements of the three 6's:

6 6 6 __ __
6 6 __ 6 __
6 6 __ __ 6
6 __ 6 6 __
6 __ 6 __ 6
6 __ __ 6 6
__ 6 6 6 __
__ 6 6 __ 6
__ 6 __ 6 6
__ __ 6 6 6

So there are 10 ways that we can arrange the 6's.

Then, we deal with the other two blanks. For each of the 10 options above, the two remaining blanks can be any digit other than 6, giving 9 options for each blank. This means that for each of the 10 options above, we have 9*9=81 total arrangements of digits for the two remaining blanks. Combining these two pieces of information, 10 arrangements of the three 6's * 81 arrangements of the remaining two blanks each = 810 favorable arrangements.

Since there are 10^5 arrangements total, the answer is B.

When you have one severely limiting factor (like needing three 6's), it can sometimes be relatively quick to brute force just that piece, then use simple combinatorics principles (like the slot method used here) for the rest. However, if you use this method, be extremely careful not to miss any options. Here, I stayed organized by locking the first two 6's and finding all of the options to place the third 6. Then, keeping the first 6 locked, I moved the second 6 by one space, locked it, and found all of the options to place the third 6. I repeated this process for all options for the second 6 before moving the first 6 one space, locking it, and repeating the process from the beginning.
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Re M26-19  [#permalink]

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New post 01 Jan 2020, 07:20
Hi Bunel,
Can you Please explain the logic of using 5c3 here?
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New post 01 Jan 2020, 08:35
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Re: M26-19   [#permalink] 01 Jan 2020, 08:35
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