GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Oct 2019, 01:45

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

M26-19

Author Message
TAGS:

Hide Tags

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

Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 58441

Show Tags

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.
_________________
Intern
Joined: 30 Apr 2017
Posts: 4

Show Tags

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.
Manager
Joined: 27 Jan 2016
Posts: 123
Schools: ISB '18
GMAT 1: 700 Q50 V34

Show Tags

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
Current Student
Joined: 22 Apr 2017
Posts: 106
Location: India
GMAT 1: 620 Q46 V30
GMAT 2: 620 Q47 V29
GMAT 3: 630 Q49 V26
GMAT 4: 690 Q48 V35
GPA: 3.7

Show Tags

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.
Intern
Joined: 15 Aug 2013
Posts: 41

Show Tags

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
Intern
Joined: 13 Jun 2015
Posts: 15

Show Tags

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
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4009

Show Tags

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
_________________
Test confidently with gmatprepnow.com
Intern
Joined: 22 May 2018
Posts: 6

Show Tags

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
Veritas Prep GMAT Instructor
Joined: 01 May 2019
Posts: 50

Show Tags

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

Go to page   Previous    1   2   [ 30 posts ]

Display posts from previous: Sort by

M26-19

Moderators: chetan2u, Bunuel