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AdjustedEbitdad
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Baker's Dozen 14 Mar 2025, 22:40
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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
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AdjustedEbitdad
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Baker's Dozen 14 Mar 2025, 22:47
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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 \(= 10*(1/10)^3 * (9/10)^2. = 810/10^5\)
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Baker's Dozen 15 Mar 2025, 02:05
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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
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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).
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Baker's Dozen 17 Mar 2025, 01:56
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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 \(= 10*(1/10)^3 * (9/10)^2. = 810/10^5\)
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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.
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Baker's Dozen 03 Sep 2025, 07:07
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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.
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Baker's Dozen 03 Sep 2025, 07:26
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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

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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.
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Baker's Dozen 16 Sep 2025, 06:42
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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.
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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.
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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 \(\sqrt{-|y|^2} = |y|=0\).
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