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Nipunh
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M26-19 23 Jul 2025, 06:00
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Hi, I am a little confused here. Help me with where I am going wrong and why.
I divided this problem into 2 parts...

in which the probability of getting 3 sixes and 2 different numbers...
9 x 8 x 5c3 = 1440

and probability of getting 3 sixes 2 of the same numbers
(9 x 1 x 5c3) / (3! x 2!) = 90
Total choices = 10^5
Adding both and dividing by total choices, I get (E)

I am not sure why the original solution has not taken these cases. Can you please explain?
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Bunuel
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M26-19 23 Jul 2025, 14:09
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Nipunh
Hi, I am a little confused here. Help me with where I am going wrong and why.
I divided this problem into 2 parts...

in which the probability of getting 3 sixes and 2 different numbers...
9 x 8 x 5c3 = 1440

and probability of getting 3 sixes 2 of the same numbers
(9 x 1 x 5c3) / (3! x 2!) = 90
Total choices = 10^5
Adding both and dividing by total choices, I get (E)

I am not sure why the original solution has not taken these cases. Can you please explain?
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9 * 8 * 5C3 = 720, not 1440.

For the second case, it's just 9 * 5C3 = 90.

Total is 810.

However, the official solution uses 9 * 9, which counts all combinations of the two remaining digits, including both same and different.
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M26-19 10 Sep 2025, 12:35
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During the test i got it wrong..but later while reviewing ...this was my approach

Total cases can be 10^5

Now...Favourable....5 spaces are there.....

The number 6 can choose any 3....So out of the 5 ...6 can choose its place in 5C3 Ways....Makes its 10...And for each place we have 9 X 9 options ..So 81 codes for each of the 10 ways....

Makes it 810....

Answer 810/10^5
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M26-19 30 Oct 2025, 02:27
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I like the solution - it’s helpful.
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M26-19 08 Jan 2026, 12:17
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think of this question
6=x and rest of the numbers are y

so first xxxyy can be arranged in 5!/3!*2!......................(1)

now probability
1*1*1*9*9/10^5...........................................(2)

Answer is (1) * (2)
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M26-19 10 Jun 2026, 06:37
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I don’t quite agree with the solution. How about when zero is at the first digit, we have to eliminate those case while calculating favourable number.
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M26-19 10 Jun 2026, 07:22
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SomyaMBA2025
I don’t quite agree with the solution. How about when zero is at the first digit, we have to eliminate those case while calculating favourable number.
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We are not looking for 5-digit numbers. Codes made with 5 digits can have first digit as 0.
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M26-19 22 Jun 2026, 00:19
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I did not quite understand the solution. If there are 5 place and 3 of there are taken by 6 then we have 2 remaining places. 5C3 is accounting for placement of 6

But what about the 2 remaining digits.

66261 and 66162

Are two different arrangements
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M26-19 22 Jun 2026, 00:31
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I was able to follow till the point where he picked the case of 2 DIFFERENT digits

So we have 3 6’s and 2 different digits

6,6,6,-,-

Now he says 2 different digits can be picked in 9C2 ways = 36 ? How and why

A more simple approach is if we have 2 blanks and we need to fill with two different digits then one can be filled with (9 options) and other ( 8 options) = 72 ways

Why are we selecting 2 digits from a set of 9 digits equalling 36 ways?

And the solution that 5C3 is used to pick 3 spots for digit 6 and remaining 2 goes to the other 2 digit, how are we handling the case of 66261 and 66162 bcz those 2 digits can also change places right?

Can someone please give a solution simple enough and detailed enough to tackle all these doubts
BrentGMATPrepNow


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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M26-19 22 Jun 2026, 02:30
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thisisakr
I did not quite understand the solution. If there are 5 place and 3 of there are taken by 6 then we have 2 remaining places. 5C3 is accounting for placement of 6

But what about the 2 remaining digits.

66261 and 66162

Are two different arrangements
Show more

The two remaining digits are accounted for by the 9*9 part.

After choosing the three positions for the 6s, the two remaining positions are still distinct positions. Each of them can be filled with any of the 9 non-6 digits.

So 66261 and 66162 are counted separately because they come from different choices for the two remaining positions.
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M26-19 24 Jul 2026, 09:54
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Hi, you are wrong here , you have calculated for the possibility of having the other 2 digits same
i.e 9*9* 5C3 but what if the other two are different numbers 9*8*5!/3! you havent added this part
Bunuel
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. \(\frac{1530}{100,000}\)


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
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M26-19 24 Jul 2026, 10:18
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abvoluptatum
Hi, you are wrong here , you have calculated for the possibility of having the other 2 digits same
i.e 9*9* 5C3 but what if the other two are different numbers 9*8*5!/3! you havent added this part

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No, you are wrong actually. The factor 9 * 9 already accounts for every possible choice of the two non-6 digits, whether they are the same or different. Please study the discussion carefully before commenting further.
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M26-19 01 Aug 2026, 05:19
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666_ _ = 9*9 - this part is understood = 81 considering 6 digit will not come now 81 will be muliplied by 5! - because digits can come anwhere divided by 3! because 6 is being repreated twice. Now why will we multiply by 2!? we donot know the remaining 2 digits are being repeated or not? There are cases where it will be repeated 66622 and there are cases where it will not be repeated 666 21 - so why do we necessarily multiply by 2!?

2nd Method - Consider 2 cases where 1) The other 2 digits will not be repeated 2) The other two digits will be repeated

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)

In this part why is answer = 36? Will it not be 72?? 666 9*8 ways for the remaining 2 digits?

I have understood the second part here

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
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M26-19 01 Aug 2026, 05:26
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Bunuel


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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Without formula, this will be 5!/ 3! righht? but this is not getting the same answer - why will we divide by 2! when we dont know the remaining digits will repeat or not?
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M26-19 01 Aug 2026, 07:03
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nikitathegreat

Without formula, this will be 5!/ 3! righht? but this is not getting the same answer - why will we divide by 2! when we dont know the remaining digits will repeat or not?
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5C3 is correct because we only need to choose which 3 of the 5 positions contain 6. Once those are chosen, the other 2 positions are automatically the non-6 positions.
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