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Re: Baker's Dozen
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11 Sep 2014, 10:49
Bunuel wrote: 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
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 digit 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.
\(P=\frac{favorable}{total}=\frac{810}{10^5}\)
Answer: B. Hi Bunuel, I was unsure how to do this problem and when I saw your explanation, I was wondering how you got ( 9 * 9 * 3C5 ) ? Not sure where the 9s came from. Thanks,



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Re: Baker's Dozen
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12 Sep 2014, 08:13
DangerPenguin wrote: Bunuel wrote: 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
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 digit 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.
\(P=\frac{favorable}{total}=\frac{810}{10^5}\)
Answer: B. Hi Bunuel, I was unsure how to do this problem and when I saw your explanation, I was wondering how you got ( 9 * 9 * 3C5 ) ? Not sure where the 9s came from. Thanks, XX666 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 5digit password unlike 5digit number can start with 0.
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Re: Baker's Dozen
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16 Nov 2014, 13:28
Bunuel wrote: 2. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), then y is NOT divisible by which of the following? A. 6^4 B. 62^2 C. 65^2 D. 15^4 E. 52^4
\(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}=(3^53^2)^2*(5^75^4)^2=3^4*(3^31)^2*5^8*(5^31)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).
Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).
Answer: E. Hi Brunel, Great set of questions! Could you please explain the logic behind how you factorised this: \(y=(3^53^2)^2*(5^75^4)^2=3^4*(3^31)^2*5^8*(5^31)^2\). I am not too clear on how you pulled out factors of 3^4 and 5^8, and still have a 3^3 and a 5^3 left in the brackets. Thanks!



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16 Nov 2014, 13:35
Madrigal wrote: Bunuel wrote: 2. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), then y is NOT divisible by which of the following? A. 6^4 B. 62^2 C. 65^2 D. 15^4 E. 52^4
\(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}=(3^53^2)^2*(5^75^4)^2=3^4*(3^31)^2*5^8*(5^31)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).
Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).
Answer: E. Hi Brunel, Great set of questions! Could you please explain the logic behind how you factorised this: \(y=(3^53^2)^2*(5^75^4)^2=3^4*(3^31)^2*5^8*(5^31)^2\). I am not too clear on how you pulled out factors of 3^4 and 5^8, and still have a 3^3 and a 5^3 left in the brackets. Thanks! \((abac)^2=(a(bc))^2=a^2*(bc)^2\). So, \((3^53^2)^2=(3^2*(3^31))^2=(3^2)^2*(3^31)^2=3^4*(3^31)^2\). Similarly \((5^75^4)^2=(5^4(5^31))^2=5^8*(5^31)^2\). Hope it's clear.
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Re: Baker's Dozen
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04 Dec 2014, 07:33
Bunuel wrote: 4. What is the smallest positive integer \(k\) such that \(126*\sqrt{k}\) is the square of a positive integer? A. 14 B. 36 C. 144 D. 196 E. 441
\(126=2*3^2*7\), so in order \(126*\sqrt{k}\) to be a square of an integer \(\sqrt{k}\) must complete the powers of 2 and 7 to even number, so the least value of \(\sqrt{k}\) must equal to 2*7=14, which makes the leas value of \(k\) equal to 14^2=196.
Answer: D. Would really appreciate if someone can explain this bit "\sqrt{k} must complete the powers of 2 and 7 to even number," Why does it have to be powers of 2*7



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Re: Baker's Dozen
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04 Dec 2014, 07:57
saadis87 wrote: Bunuel wrote: 4. What is the smallest positive integer \(k\) such that \(126*\sqrt{k}\) is the square of a positive integer? A. 14 B. 36 C. 144 D. 196 E. 441
\(126=2*3^2*7\), so in order \(126*\sqrt{k}\) to be a square of an integer \(\sqrt{k}\) must complete the powers of 2 and 7 to even number, so the least value of \(\sqrt{k}\) must equal to 2*7=14, which makes the leas value of \(k\) equal to 14^2=196.
Answer: D. Would really appreciate if someone can explain this bit "\sqrt{k} must complete the powers of 2 and 7 to even number," Why does it have to be powers of 2*7 Let me ask you: by what positive integer we should multiply \(126=2*3^2*7\) for the product to be the square of an integer?
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Re: Baker's Dozen
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21 Dec 2014, 23:44
Bunuel wrote: 6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool? A. \(\frac{yz}{x+y+z}\)
B. \(\frac{yz}{yz+xzxy}\)
C. \(\frac{yz}{yz+xz+xy}\)
D. \(\frac{xyz}{yz+xzxy}\)
E. \(\frac{yz+xzxy}{yz}\)
With pumps A and B both running and the drain unstopped the pool will be filled in a rate \(\frac{1}{x}+\frac{1}{y}\frac{1}{z}=\frac{yz+xzxy}{xyz}\) pool/hour. So, the pool will be filled in \(\frac{xyz}{yz+xzxy}\) hours (time is reciprocal of rate).
In \(\frac{xyz}{yz+xzxy}\) hours A will pump \(\frac{1}{x}*\frac{xyz}{yz+xzxy}=\frac{yz}{yz+xzxy}\) amount of the water into the pool.
Answer: B. I did not get why 1/x is multiplied. Would you mind terribly, to explain? Thanks



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22 Dec 2014, 03:19
deeuk wrote: Bunuel wrote: 6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool? A. \(\frac{yz}{x+y+z}\)
B. \(\frac{yz}{yz+xzxy}\)
C. \(\frac{yz}{yz+xz+xy}\)
D. \(\frac{xyz}{yz+xzxy}\)
E. \(\frac{yz+xzxy}{yz}\)
With pumps A and B both running and the drain unstopped the pool will be filled in a rate \(\frac{1}{x}+\frac{1}{y}\frac{1}{z}=\frac{yz+xzxy}{xyz}\) pool/hour. So, the pool will be filled in \(\frac{xyz}{yz+xzxy}\) hours (time is reciprocal of rate).
In \(\frac{xyz}{yz+xzxy}\) hours A will pump \(\frac{1}{x}*\frac{xyz}{yz+xzxy}=\frac{yz}{yz+xzxy}\) amount of the water into the pool.
Answer: B. I did not get why 1/x is multiplied. Would you mind terribly, to explain? Thanks The question asks to find the amount of water in terms of the fraction of the pool which pump A pumped into the pool. We know that the pool will be filled in \(\frac{xyz}{yz+xzxy}\) hours and the rate of A is 1/x pool/hour. The job done by A in that time is (job) = (time)*(rate) = \(\frac{ 1}{x}*\frac{xyz}{yz+xzxy}\). Hope it's clear.
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Re: Baker's Dozen
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22 Dec 2014, 21:01
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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Re: Baker's Dozen
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23 Dec 2014, 22:31
VeritasPrepKarishma wrote: 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?
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. Can I use permutation for the 2 non6 digits? 5C3 is for choosing 3 spots to place 6s in 5 spots, for the remaining 2 spots one of 9 digits are used. Can this be expressed with '9C2' and something?



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Re: Baker's Dozen
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25 Dec 2014, 22:56
deeuk wrote: VeritasPrepKarishma wrote: 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?
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. Can I use permutation for the 2 non6 digits? 5C3 is for choosing 3 spots to place 6s in 5 spots, for the remaining 2 spots one of 9 digits are used. Can this be expressed with '9C2' and something? No. With 9C2, you are selecting 2 digits out of 9. There are two problems with it: 1. You are assuming that the two digits are distinct. Nothing stops the password from being 66622. 2. You are not considering that the two digits are for two distinct places. 66623 is different from 66632 but with 9C2, you are assuming that it is the same password.
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Re: Baker's Dozen
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19 Feb 2015, 18:22
Bunuel wrote: 4. What is the smallest positive integer \(k\) such that \(126*\sqrt{k}\) is the square of a positive integer? A. 14 B. 36 C. 144 D. 196 E. 441
\(126=2*3^2*7\), so in order \(126*\sqrt{k}\) to be a square of an integer \(\sqrt{k}\) must complete the powers of 2 and 7 to even number, so the least value of \(\sqrt{k}\) must equal to 2*7=14, which makes the leas value of \(k\) equal to 14^2=196.
Answer: D. Hi Bunuel, Thanks for your great workouts. I have a question on this one though. If K= B (36), the number would become a square of a positive integer as well. For that matter any perfect square number will do  say 4?! \(126*\sqrt{36}\) = 756. Why can't B be an answer. Your comment is highly appreciated.



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20 Feb 2015, 03:27
falewman wrote: Bunuel wrote: 4. What is the smallest positive integer \(k\) such that \(126*\sqrt{k}\) is the square of a positive integer? A. 14 B. 36 C. 144 D. 196 E. 441
\(126=2*3^2*7\), so in order \(126*\sqrt{k}\) to be a square of an integer \(\sqrt{k}\) must complete the powers of 2 and 7 to even number, so the least value of \(\sqrt{k}\) must equal to 2*7=14, which makes the leas value of \(k\) equal to 14^2=196.
Answer: D. Hi Bunuel, Thanks for your great workouts. I have a question on this one though. If K= B (36), the number would become a square of a positive integer as well. For that matter any perfect square number will do  say 4?! \(126*\sqrt{36}\) = 756. Why can't B be an answer. Your comment is highly appreciated. \(126*\sqrt{k}\) must be a prefect square. If k=36, \(126*\sqrt{k}=756\), which is not a perfect square.
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26 Nov 2015, 11:47
Bunuel wrote: 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. 1y C. 1y D. y1 E. xy
Note that \(\sqrt{a^2}=a\). Next, since \(x<0\) and \(y<0\) then \(x=x\) and \(y=y\).
So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\)
Answer: D. How is it that the absolute value of y is still a negative (i.e y = (y)) under the root? My understanding is an absolute value can never be negative
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26 Nov 2015, 11:58
redfield wrote: Bunuel wrote: 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. 1y C. 1y D. y1 E. xy
Note that \(\sqrt{a^2}=a\). Next, since \(x<0\) and \(y<0\) then \(x=x\) and \(y=y\).
So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\)
Answer: D. How is it that the absolute value of y is still a negative (i.e y = (y)) under the root? My understanding is an absolute value can never be negative You are correct that an absolute value of any number is >0 but HOW does the absolute of any number give us a positive value? For any number \(\geq\) 0 , y = y but for numbers y<0 , y = y (this is how you make the absolute value of a negative number as a positive quantity). Another thing to note here is that for GMAT quant, \(\sqrt {x}\) , x MUST be \(\geq\)0. So for making \((y)*y \geq 0\), y = y (as it is given that y <0)



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26 Nov 2015, 12:06
Engr2012 wrote: You are correct that an absolute value of any number is >0 but HOW does the absolute of any number give us a positive value?
For any number \(\geq\) 0 , y = y but
for numbers y<0 , y = y (this is how you make the absolute value of a negative number as a positive quantity).
Another thing to note here is that for GMAT quant, \(\sqrt {x}\) , x MUST be \(\geq\)0.
So for making \((y)*y \geq 0\), y = y (as it is given that y <0)
Sorry not quite following entirely; on the GMAT if I see an absolute power of a variable less than zero (i.e negative) it can still be negative? Or is that the case only where the square root is involved? So here's where I'm coming from on this, if absolute value is just the distance from zero, even a negative number < 0 such as 2 would still be 2 away from 0? Thank you for your patience on my math skills and with your answers Engr2012, I know I've been asking rudimentary questions all over this forum the last few hours!
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26 Nov 2015, 12:09
redfield wrote: Engr2012 wrote: You are correct that an absolute value of any number is >0 but HOW does the absolute of any number give us a positive value?
For any number \(\geq\) 0 , y = y but
for numbers y<0 , y = y (this is how you make the absolute value of a negative number as a positive quantity).
Another thing to note here is that for GMAT quant, \(\sqrt {x}\) , x MUST be \(\geq\)0.
So for making \((y)*y \geq 0\), y = y (as it is given that y <0)
Sorry not quite following entirely; on the GMAT if I see an absolute power of a number less than zero (i.e negative)? So here's where I'm coming from on this, if absolute value is just the distance from zero, even a negative number < 0 such as 2 would still be 2 away from 0? Thank you for your patience on my math skills and with your answers Engr2012, I know I've been asking rudimentary questions all over this forum the last few hours! Both of us are saying the same thing. Consider the same example that you are using. When x = 2, x = +2 . How did you come about getting this value of '2' from '2'? You multiplied 2 (the actual value of x ) by 1 to get 1*2 = +2. This is what absolute value of a negative number does. When you have y<0, y =  (ngeative number ) = + number. Absolute values will ALWAYS be \(\geq 0\) as it is the "distance" of the number from 'zero' on the number line. Hope this helps.



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Re: Baker's Dozen
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11 Jan 2016, 06:33
Bunuel wrote: 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
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 digit 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.
\(P=\frac{favorable}{total}=\frac{810}{10^5}\)
Answer: B. Hi Bunuel, I have a doubt in 5C3 part in this problem. I solved something like this: Code will be something like this 6,6,6,A,B this is like arranging 5 letters where 3 are exactly similar. It would give: 5! / 3! which is not equal to 5C3 . Therefore I am getting a wrong answer. Please correct my mistake (and why we are using Combination here while it looks like a case of Permutation). Thanks in advance. SR



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Re: Baker's Dozen
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11 Jan 2016, 08:47
solitaryreaper wrote: Bunuel wrote: 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
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 digit 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.
\(P=\frac{favorable}{total}=\frac{810}{10^5}\)
Answer: B. Hi solitaryreaper, there are various Queries on this Q that why don't we use permutations and use combinations.. that is why 5!/3!2! instead of 5!/3!.. firstly, yes the Question is of permutation, but we still require to use combination formula.. WHY? we have choosen three 6 digits and 2 digits as any of the remaining three.. answer is 9*9*5C3.... and not 9*9*5P3 because the permutations will be ok till the time we have two separate digits... and it will be combinations the moment the other two digits are same.... so how can we use permutations even when digits are same, and that is what we are doing when we take other two digits as 9*9..then how do we arrive at 9*9*5!/3!2!.. there are two ways we can take the remaining two digits.. 1) three 6s and both other digits are different.. aaabc b and c can be chosen out of 9 digits ways = 9C2*5!/3!=(9*8)/2! * 5!/3!...2)three 6s and both other digits are same.. aaabb.. b can be selected in 9 ways ways = 9*5!/3!2!.. total ways = add the two=9*8*5/3!2! + 9*5!/3!2!= 9*5!/3!2!* (8+1)=9*9*5!/3!2!=9*9*5C3.. I hope this clears the air around the use of permutation or combination in this Q..
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Re: Baker's Dozen
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12 Jan 2016, 00:27
solitaryreaper wrote: Bunuel wrote: 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
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 digit 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.
\(P=\frac{favorable}{total}=\frac{810}{10^5}\)
Answer: B. Hi Bunuel, I have a doubt in 5C3 part in this problem. I solved something like this: Code will be something like this 6,6,6,A,B this is like arranging 5 letters where 3 are exactly similar. It would give: 5! / 3! which is not equal to 5C3 . Therefore I am getting a wrong answer. Please correct my mistake (and why we are using Combination here while it looks like a case of Permutation). Thanks in advance. Now I am having a hard time to grasp how combination is giving the same answer. Can you please help me with the combination approach(how the problem translates to combination which is equivalent to the one the we got via permutation) ? Moreover , how to decide when to use Permutation or Combination (and one that would lead to the solution in lesser time) SR Responding to a pm: This is a question that haunts many  how do I know when to use permutation and when to use combination? There is a simple solution  never use the permutation formula. The permutation formula finds very little direct use but leads to too many complications. Always think in terms of selecting and arranging. For selecting r distinct elements out of n distinct elements, use nCr formula. Then arrange depending on whether the r elements selected need to be all distinct (r!) or some need to be same (a!/b!*c!) etc Here, out of 9 digits you can select 2 in 9C2 ways and get a case such as 666AB. This will be arranged in 5!/3! ways. Or you can select 1 digit out of 9 in 9C1 ways and get a case such as 666AA. This will be arranged in 5!/2!*3! ways. 9C2*5!/3! + 9C1*5!/2!*3! = 9*5!/3! * 9/2 = 810 Probability = 810/10^5 Note that you are using the Combinations formula only in this method too. The other method using the combination formula just shows a different way of thinking. You have 5 spots: _____ _____ _____ _____ _____ You choose any two spots out of these 5 in 5C2 ways. For the first spot you choose, select a digit in 9 ways. For the second spot, select a digit in 9 ways. In all remaining spots, just put 6. This 5C2*9*9 directly gives you the total number of ways. Note that 5C2 is the same as 5C3 (either you choose 2 spots for non 6 digits or you choose 3 spots for 6).
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Re: Baker's Dozen &nbs
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