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# M26-19

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Intern
Joined: 24 Mar 2016
Posts: 4
GMAT 1: 710 Q50 V37

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16 Jun 2017, 22:51
1
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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17 Jun 2017, 04:24
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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29 Sep 2017, 23:31
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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03 Nov 2017, 02:15
4
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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22 Nov 2017, 00:59
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.

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

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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Joined: 15 Aug 2013
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10 Dec 2017, 10:12
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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03 Aug 2018, 04:37
Hi

Would it be conceptually okay to solve this question using Binomial? 5C3*(1/10)^3*(9/10)^2
1/10 being the success of there being a 6 in the code and 9/10 being the converse?

Thanks
SN
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Joined: 12 Sep 2015
Posts: 4009

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30 Jun 2019, 10:36
Top Contributor
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

Cheers,
Brent
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30 Jul 2019, 08:29
To have only three digits 6, the possibilities for the other digits, drops from 10 digits to 9 digits.

then, 9 x 9 x 1 x 1 x 1 = 81 possible combinations of digits. When we consider 9 x 9, the options A X A, A X B and B X A are already accounted.

Then, when we think about the order, we have to consider the other 2 digits as the same to not double count it. A X A

So, 5! / (3!x2!) -> 3! for the repeated 6 and 2! for the digits that we already accounted the order by multipling 9 x 9

Then,
9 x 9 x 1 x 1 x 1 = 81
5! / (3!x2!) = 10
81*10 =810
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Joined: 01 May 2019
Posts: 50

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30 Jul 2019, 10:28
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.
Re: M26-19   [#permalink] 30 Jul 2019, 10:28

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# M26-19

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