Last visit was: 25 Apr 2026, 10:06 It is currently 25 Apr 2026, 10:06
Home
Messages
Tests
Schools
Events
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2   3   4   5 
User avatar
AdjustedEbitdad
User avatar
Joined: 22 Apr 2022
Last visit: 27 Mar 2026
Posts: 24
Own Kudos:
6
Given Kudos: 120
Location: India
GMAT Focus 1: 665 Q82 V84 DI83
GMAT 1: 710 Q50 V35
Products:
My Reviews
GMAT Focus 1: 665 Q82 V84 DI83
GMAT 1: 710 Q50 V35
Posts: 24
Kudos: 6
Baker's Dozen 14 Mar 2025, 22:40
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunuel can you please help know why the Numerator isn't 9*9*1*1*1*5!/3!? Out of 5, 3 are identical.

I understand there can be cases where the other two are also identical. However, I don't understand how 5C3 handles those cases or how it positions the other two.
Bunuel
SOLUTIONS:

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

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

The total number of 5-digit codes is \(10^5\). It's important to note that it's not \(9*10^4\), as the first digit can be zero in a password.

The number of passwords with three digits as 6 can be calculated as \(9*9*C^3_5 = 810\). We have 9 choices for each of the two remaining digits (not 6), resulting in \(9*9\). The term \(C^3_5\) represents the number of ways to choose the positions for the 6s among the five digits (essentially deciding which three out of ***** will be 6s).

The probability is therefore, \(P=\frac{favorable}{total}=\frac{810}{10^5}\).


Answer: B
Show more
User avatar
AdjustedEbitdad
User avatar
Joined: 22 Apr 2022
Last visit: 27 Mar 2026
Posts: 24
Own Kudos:
6
 [1]
Given Kudos: 120
Location: India
GMAT Focus 1: 665 Q82 V84 DI83
GMAT 1: 710 Q50 V35
Products:
My Reviews
GMAT Focus 1: 665 Q82 V84 DI83
GMAT 1: 710 Q50 V35
Posts: 24
Kudos: 6
 [1]
Baker's Dozen 14 Mar 2025, 22:47
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Hi KarishmaB

I have a small doubt here. Aren't there cases where the numbers are 6,6,6, N1, and N2, and for them, the arrangement should be 5!/3!?
KarishmaB
Quote:

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

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000


I see that (9*9*1*1*1)5C3 is a popular answer method. I agree with this approach, but i was wondering, since permutation is choosing and arranging, 'why not permutation?'. I must be wrong here and i'd really like to know where/ how.
I did read all the other explanations. The repetition approach(5!/3!) looked good too. But i did not understand why a few of em used the probability approach (10*1/10*1/10....).
What would your process for this be?

Responding to a pm:

Permutation is choosing and arranging - correct! But you need to know what you are arranging. Permutation (written as nPr) implies choose r things out of n and then arrange the r things among themselves. Say you select 3 chocolates out of 5 to give as prizes in a game and then arrange the 3 selected chocolates in three spots - first, second and third. The one who stands first, gets the first chocolate and so on. So if you use permutation here, you are selecting 3 spots where you will put 6 and then arranging them in 3! ways. What we actually need to do is select 3 spots out of 5 and put 6 in them. We have to then select 2 digits for the leftover 2 spots, wherever they are!

Hence we select the three spots in 5C3 ways and then choose a digit for the remaining two places in 9*9 ways.

Probability approach:
There are 5 spots ___ ___ ___ ___ ___

What is the probability that the number looks like 6N6N6? (1/10)*(9/10)*(1/10)*(9/10)*(1/10)
N is any number other than 6.

What is the probability that the number looks like N66N6? (9/10)*(1/10)*(1/10)*(9/10)*(1/10)

Similarly there are many other ways of making the password with three 6s. You need to add all some probabilities. The probability in each case will be \((1/10)^3 * (9/10)^2\). How many such distinct cases will be there? It depends on the number of ways in which you can arrange the 6s and Ns = 5!/3!*2! = 10
Total probability \(= 10*(1/10)^3 * (9/10)^2. = 810/10^5\)
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 25 Apr 2026
Posts: 109,827
Own Kudos:
811,227
Given Kudos: 105,878
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,827
Kudos: 811,227
Baker's Dozen 15 Mar 2025, 02:05
Kudos
Add Kudos
Bookmarks
Bookmark this Post
AdjustedEbitdad
Hi Bunuel can you please help know why the Numerator isn't 9*9*1*1*1*5!/3!? Out of 5, 3 are identical.

I understand there can be cases where the other two are also identical. However, I don't understand how 5C3 handles those cases or how it positions the other two.
Bunuel
SOLUTIONS:

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

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

The total number of 5-digit codes is \(10^5\). It's important to note that it's not \(9*10^4\), as the first digit can be zero in a password.

The number of passwords with three digits as 6 can be calculated as \(9*9*C^3_5 = 810\). We have 9 choices for each of the two remaining digits (not 6), resulting in \(9*9\). The term \(C^3_5\) represents the number of ways to choose the positions for the 6s among the five digits (essentially deciding which three out of ***** will be 6s).

The probability is therefore, \(P=\frac{favorable}{total}=\frac{810}{10^5}\).


Answer: B
Show more
Here is the logic behind it:

The term \(C^3_5\) represents the number of ways to choose the positions for the 6s among the five digits (essentially deciding which three out of ***** will be 6s).
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 23 Apr 2026
Posts: 16,442
Own Kudos:
79,411
Given Kudos: 485
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,442
Kudos: 79,411
Baker's Dozen 17 Mar 2025, 01:56
Kudos
Add Kudos
Bookmarks
Bookmark this Post
AdjustedEbitdad
Hi KarishmaB

I have a small doubt here. Aren't there cases where the numbers are 6,6,6, N1, and N2, and for them, the arrangement should be 5!/3!?
KarishmaB
Quote:

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

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000


I see that (9*9*1*1*1)5C3 is a popular answer method. I agree with this approach, but i was wondering, since permutation is choosing and arranging, 'why not permutation?'. I must be wrong here and i'd really like to know where/ how.
I did read all the other explanations. The repetition approach(5!/3!) looked good too. But i did not understand why a few of em used the probability approach (10*1/10*1/10....).
What would your process for this be?

Responding to a pm:

Permutation is choosing and arranging - correct! But you need to know what you are arranging. Permutation (written as nPr) implies choose r things out of n and then arrange the r things among themselves. Say you select 3 chocolates out of 5 to give as prizes in a game and then arrange the 3 selected chocolates in three spots - first, second and third. The one who stands first, gets the first chocolate and so on. So if you use permutation here, you are selecting 3 spots where you will put 6 and then arranging them in 3! ways. What we actually need to do is select 3 spots out of 5 and put 6 in them. We have to then select 2 digits for the leftover 2 spots, wherever they are!

Hence we select the three spots in 5C3 ways and then choose a digit for the remaining two places in 9*9 ways.

Probability approach:
There are 5 spots ___ ___ ___ ___ ___

What is the probability that the number looks like 6N6N6? (1/10)*(9/10)*(1/10)*(9/10)*(1/10)
N is any number other than 6.

What is the probability that the number looks like N66N6? (9/10)*(1/10)*(1/10)*(9/10)*(1/10)

Similarly there are many other ways of making the password with three 6s. You need to add all some probabilities. The probability in each case will be \((1/10)^3 * (9/10)^2\). How many such distinct cases will be there? It depends on the number of ways in which you can arrange the 6s and Ns = 5!/3!*2! = 10
Total probability \(= 10*(1/10)^3 * (9/10)^2. = 810/10^5\)
Show more

Yes, there will be and they are counted when we use:
\((1/10)^3 * (9/10)^2\)

1/10 is for the three 6s and 9/10 are the number of ways of selecting non 6 number. It could be anything. So both Ns could be same or different.
This is similar to selecting 3 of the 5 spots in 5C3 = 5!/3!*2! ways and putting the three 6s there with probability 1/10. In the remaining 2 spots, we put non 6s with the probability 9/10 each.
User avatar
Raome
User avatar
Joined: 21 Apr 2025
Last visit: 16 Feb 2026
Posts: 114
Own Kudos:
31
Given Kudos: 84
Location: India
Posts: 114
Kudos: 31
Baker's Dozen 03 Sep 2025, 07:07
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello,

Thank you so much for this collection, Bunuel. I appreciate it!
I have a question here:

why can't i do the following:-

(8!)^6 [(8!)^4 - 1]/ (8!)^3 [ (8!)^2 -1] = (8!)^3 [ (8!)^2] (since 8!^3 is much larger than 1) = (8!)^5
Bunuel
13. If \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}-39\)?
A. 0
B. 6
C. 7
D. 12
E. 14

Apply \(a^2-b^2=(a-b)(a+b)\): \(x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{((8!)^{5}-(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}-(8!)^3}=(8!)^{5}+(8!)^3\).

Next, \(\frac{x}{(8!)^3}-39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}-39=(8!)^2+1-39=(8!)^2-38\).

Now, since \(8!\) has 2 and 5 as its multiples, then it will have 0 as the units digit, so \((8!)^2\) will have two zeros in the end, which means that \((8!)^2-38\) will have 00-38=62 as the last digits: 6*2=12.

Answer: D.
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 25 Apr 2026
Posts: 109,827
Own Kudos:
811,227
Given Kudos: 105,878
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,827
Kudos: 811,227
Baker's Dozen 03 Sep 2025, 07:26
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Raome
Hello,

Thank you so much for this collection, Bunuel. I appreciate it!
I have a question here:

why can't i do the following:-

(8!)^6 [(8!)^4 - 1]/ (8!)^3 [ (8!)^2 -1] = (8!)^3 [ (8!)^2] (since 8!^3 is much larger than 1) = (8!)^5

Show more

We’re asked for the product of the tens and units digits of the expression, so an approximation won’t work here. We need the exact value, not an estimate, in order to determine the correct digits.
User avatar
hanbyeoli
User avatar
Joined: 16 Aug 2024
Last visit: 30 Dec 2025
Posts: 6
Given Kudos: 29
Location: India
Posts: 6
Kudos: 0
Baker's Dozen 16 Sep 2025, 06:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunuel,

The solution is clear and I have solved it correctly, but I just have a doubt here about something which I have noticed in other solutions as well.
For a value like this- \sqrt{(-(y^2))} = i*|y| but in gmat, we do not take iota and write the expression as -|y|. Why do we do this? Is this mathematically correct?

Bunuel
9. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}\)?
A. 1+y
B. 1-y
C. -1-y
D. y-1
E. x-y

Note that \(\sqrt{a^2}=|a|\). Next, since \(x<0\) and \(y<0\) then \(|x|=-x\) and \(|y|=-y\).

So, \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y\)

Answer: D.
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 25 Apr 2026
Posts: 109,827
Own Kudos:
811,227
Given Kudos: 105,878
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,827
Kudos: 811,227
Baker's Dozen Updated on: 16 Sep 2025, 08:10
Originally posted by Bunuel on 16 Sep 2025, 07:43.
Last edited by Bunuel on 16 Sep 2025, 08:10, edited 3 times in total.
Kudos
Add Kudos
Bookmarks
Bookmark this Post
hanbyeoli
Hi Bunuel,

The solution is clear and I have solved it correctly, but I just have a doubt here about something which I have noticed in other solutions as well.
For a value like this- \sqrt{(-(y^2))} = i*|y| but in gmat, we do not take iota and write the expression as -|y|. Why do we do this? Is this mathematically correct?


Show more

\(\sqrt{-|y|^2} = i|y|\). Nowhere will you find that it generally equals |y|. GMAT does not have its own math. The point is, all numbers on the GMAT are real numbers; GMAT does not deal with complex numbers. So for even roots on the GMAT to be defined, the expression under them must be nonnegative. So, for \(\sqrt{-|y|^2} =\) to be defined on the GMAT y must be 0, in this case \(\sqrt{-|y|^2} = |y|=0\).
User avatar
Gaurav1144
User avatar
Joined: 02 Nov 2023
Last visit: 24 Apr 2026
Posts: 2
Own Kudos:
1
Given Kudos: 93
Posts: 2
Kudos: 1
Baker's Dozen 21 Dec 2025, 03:11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel


That's a tough 700+ problem.

We are interested 666XX numbers. Now, two this can be arranged in 10 ways;
666XX;
66X6X;
6X66X;
X666X;

66XX6;
6X6X6;
X66X6;

6XX66;
XX666;

X6X66.

So, basically with \(C^3_5=10\) we are choosing which 3 places out of 5 will be occupied by X's. Or we could do the other way around: \(C^2_510\) choosing which 2 places out of 5 will be occupied by 6's.

Next, each X can take any value from 0 to 9, inclusive but 6, so total of 9 values. Notice that the first digit can be zero here since the 5-digit password unlike 5-digit number can start with 0.

Total \(9*9*C^3_5=810\).

Hope it's clear.
Show more
for q1. I got the above explanation, i.e., 5C3 is selecting 3 places for digit '6' and then other 2 places would have 9 choices each so 9^2 , thus 9*9* 5C3.

Can you explain though, why the following yields a wrong result?

I tried to separate the numerator into 2 cases:
1. 3 '6's & 2 distinct digits: 9*8 for selecting other 2 digits and their arrangement in 5!/3! ways = 72*20 = 1440
2. 3 '6's & 2 same digits: 9 for selecting 2 same digits and their arrangement in 5!/(3!*2!) ways = 9*10 = 90

giving total of 1530 ways out of 10^5
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 25 Apr 2026
Posts: 109,827
Own Kudos:
811,227
 [1]
Given Kudos: 105,878
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,827
Kudos: 811,227
 [1]
Baker's Dozen Updated on: 21 Dec 2025, 05:30
Originally posted by Bunuel on 21 Dec 2025, 04:30.
Last edited by Bunuel on 21 Dec 2025, 05:30, edited 2 times in total.
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Gaurav1144

for q1. I got the above explanation, i.e., 5C3 is selecting 3 places for digit '6' and then other 2 places would have 9 choices each so 9^2 , thus 9*9* 5C3.

Can you explain though, why the following yields a wrong result?

I tried to separate the numerator into 2 cases:
1. 3 '6's & 2 distinct digits: 9*8 for selecting other 2 digits and their arrangement in 5!/3! ways = 72*20 = 1440
2. 3 '6's & 2 same digits: 9 for selecting 2 same digits and their arrangement in 5!/(3!*2!) ways = 9*10 = 90

giving total of 1530 ways out of 10^5
Show more
Your case split is fine. The error is in case 1: 9 * 8 treats the two non 6 digits as an ordered pair, but 5!/3! already accounts for both orders of those two digits. So you are double counting. So case 1 should be:

9C2 * 5!/3! = 36 * 20 = 720. Here 9C2 gives unordered pairs of distinct digits out of 9, and then 5!/3! arranges those two digits and the three 6s in all distinct ways.

Then case 2 is:

9 * 5!/(3!2!) = 9 * 10 = 90

Total = 720 + 90 = 810.
User avatar
Gaurav1144
User avatar
Joined: 02 Nov 2023
Last visit: 24 Apr 2026
Posts: 2
Own Kudos:
1
 [1]
Given Kudos: 93
Posts: 2
Kudos: 1
 [1]
Baker's Dozen Updated on: 21 Dec 2025, 05:31
Originally posted by Gaurav1144 on 21 Dec 2025, 05:25.
Last edited by Bunuel on 21 Dec 2025, 05:31, edited 1 time in total.
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Got it, thanks alot..
Bunuel

Your case split is fine. The error is in case 1: 9 * 8 treats the two non 6 digits as an ordered pair, but 5!/3! already accounts for both orders of those two digits. So you are double counting. So case 1 should be:

9C2 * 5!/3! = 36 * 20 = 720. Here 9C2 gives unordered pairs of distinct digits out of 9, and then 5!/3! arranges those two digits and the three 6s in all distinct ways.

Then case 2 is:

9 * 5!/(3!2!) = 9 * 10 = 90

Total = 720 + 90 = 810.
Show more
   1   2   3   4   5 
Moderators:
Bunuel
Math Expert
109827 posts
Krunaal
Tuck School Moderator
852 posts