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Baker's Dozen

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I'm posting the next set of medium/hard PS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!

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

Solution: baker-s-dozen-128782-20.html#p1057502

2. If \(y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}\), 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

Solution: baker-s-dozen-128782-20.html#p1057503

3. For the past k days the average (arithmetic mean) cupcakes per day that Liv baked was 55. Today Bibi joined and together with Liv they baked 100 cupcakes, which raises the average to 60 cupcakes per day. What is the value of k?
A. 6
B. 8
C. 9
D. 10
E. 12

Solution: baker-s-dozen-128782-20.html#p1057504

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

Solution: baker-s-dozen-128782-20.html#p1057505

5. There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble to remain in the jar?
A. 460
B. 490
C. 493
D. 455
E. 445

Solution: baker-s-dozen-128782-20.html#p1057507

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. \(\frac{yz}{yz+xz-xy}\)

C. \(\frac{yz}{yz+xz+xy}\)

D. \(\frac{xyz}{yz+xz-xy}\)

E. \(\frac{yz+xz-xy}{yz}\)

Solution: baker-s-dozen-128782-20.html#p1057508

7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive?
A. $60,000
B. $90,000
C. $100,000
D. $120,000
E. $180,000

Solution: baker-s-dozen-128782-20.html#p1057509

8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?
A. -165
B. -175
C. -195
D. -205
E. -215

Solution: baker-s-dozen-128782-20.html#p1057512

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

Solution: baker-s-dozen-128782-20.html#p1057514

10. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x-2y?
A. 7
B. 11
C. 13
D. 17
E. 19

Solution: baker-s-dozen-128782-20.html#p1057515

11. In an infinite sequence 1, 3, 9, 27, ... each term after the first is three times the previous term. What is the difference between the sum of 13th and 15th terms and the sum of 12th and 14th terms of the sequence?
A. 10*3^11
B. 20*3^11
C. 10*3^12
D. 40*3^11
E. 20*3^12

Solution: baker-s-dozen-128782-40.html#p1057517

12. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z?
A. 12
B. 20
C. 24
D. 29
E. 33

Solution: baker-s-dozen-128782-40.html#p1057519

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

Solution: baker-s-dozen-128782-40.html#p1057520
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New post 22 Dec 2014, 08:38
2. If y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-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

Answer: E
Y = 3^4*(3-1)^2*(3^2+3+1)^2 * 5^8*(5-1)^4*(5^2+5+1)^4 = 2^10*3^4*5^8*13^2*31^4
where
6^4=2^4*3^4
62^2=2^2*31^2
65^2=5^2*13^2
15^4=3^4*5^4
52^4=2^8*13^4

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New post 22 Dec 2014, 08:56
3. For the past k days the average (arithmetic mean) cupcakes per day that Liv baked was 55. Today Bibi joined and together with Liv they baked 100 cupcakes, which raises the average to 60 cupcakes per day. What is the value of k?
A. 6
B. 8
C. 9
D. 10
E. 12

Answer: B
55K+100=60*(K+1)

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New post 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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I get it now, and more...Kudos

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New post 23 Dec 2014, 01:11
Three 6's can take any 3 places out of 5 in 5C3 = 10 ways.
The other 2 blank places can be filled by any 2 digits other than 6 in 9*9 = 81 ways
Hence, reqd prob = (10 * 81)/10^5 = 810/100000

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New post 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 non-6 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 [#permalink]

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New post 25 Dec 2014, 04:53
Hey, I didn't understand the solution for the first one. Could you please explain a bit?

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New post 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 non-6 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 [#permalink]

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New post 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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New post 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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Collection of Questions:
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New post 21 Aug 2015, 19:01
I took a different approach and i am getting answer. Can you please tell me where i got it wrong

Approach ...

Scenario - 1
first 3 digits 6 and 2 digits different 666 a b

a and b can be selected in 9 * 9 ways. And this can be arranged in 9 *9 * 5! / 3! = 1440

scenario 2

first 3 digits 6 and 2 digits same 666 aa. This can be arranged is 9 * 5! /(3!* 2! ) = 90

Giving a total of 1530 . Giving me option E. Can you please tell me what is wrong in my approach

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.

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New post 26 Sep 2015, 10:14
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.


Bunuel, why dont we use the permutation (5!/3!) for the number of arrangement instead of 5C3?

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New post 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. 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.


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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New post 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. 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.


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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New post 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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Re: Baker's Dozen [#permalink]

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New post 26 Nov 2015, 12:09
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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 [#permalink]

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New post 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 [#permalink]

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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 [#permalink]

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New post 11 Jan 2016, 11:16
chetan2u wrote:
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..



Hi chetan2u

That's an awesome reply. :-D :) I was having a very hard time getting to the right solution using the permutation. Because as soon as I saw arrangement , I just went on with permutation. This clears a lot of air.thanks a lot !!!

But I can see that everyone has used Combination to solve this problem. And it is evident that solution is relatively easy using the combination approach.
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)

Thanks in advance !

Regards
SR

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Re: Baker's Dozen   [#permalink] 11 Jan 2016, 11:16

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