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# Baker's Dozen

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Joined: 27 Jul 2011
Posts: 162

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19 Apr 2013, 00:20
Bunuel wrote:
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!

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

Actually need help from the moderators to know where my approach is going wrong:

it is it is told at least 1 red , at least one blue will be there.
So i am leaving 1 red , 1 blue=> Now i have 6 Red , 4 Blue ..I have to select 8 marbels...
So it means i have to select 8 marbels out of 10 marbels (colors don't matter)
= $$C^10_8$$ = $$C^10_2$$ = 45

I know i am not getting the ans...but what's wrong with this I m not able to spot...
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21 Apr 2013, 08:19
1
sujit2k7 wrote:
Bunuel wrote:
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!

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

Actually need help from the moderators to know where my approach is going wrong:

it is it is told at least 1 red , at least one blue will be there.
So i am leaving 1 red , 1 blue=> Now i have 6 Red , 4 Blue ..I have to select 8 marbels...
So it means i have to select 8 marbels out of 10 marbels (colors don't matter)
= $$C^10_8$$ = $$C^10_2$$ = 45

I know i am not getting the ans...but what's wrong with this I m not able to spot...

Responding to a pm:

From the options, it is obvious that the marbles are considered unique i.e. the red marbles are all different, say they are numbered, and the blue marbles are all different. The problem with your approach is that when you leave 1 red and 1 blue, you don't provide for this.
Also, once you account for the uniqueness, there will be double counting - when you leave Red1 and Blue1 and then choose 8 out of the remaining 10 and say you are left with Red2 and Red3 too, this will be counted as different from when you leave Red2 and Blue1 and then choose 8 out of the remaining 10 and you are left with Red1 and Red3. These two cases are actually the same.

Rule of thumb is that 'at least one' cases are best done using the complement approach which is 'Total - None' (as shown by Bunuel in the solution).
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11 May 2013, 19:25
Bunuel wrote:
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

Given $$x=qy+3$$, where $$q$$ is a quotient, an integer $$\geq0$$. Which means that the least value of $$x$$ is when $$q=0$$, in that case $$x=3$$. This basically means that $$x$$ is less than $$y$$. For example 3 divided by 4 yields remainder of 3.

Thus we have that:
$$x$$ is divided by $$y$$ the remainder is 3 --> minimum value of $$x$$ is 3;
$$y$$ is divided by $$z$$ the remainder is 8 --> minimum value of $$y$$ is 8 and minimum value of $$z$$ is one more than 8, so 9 (8 divided by 9 yields the remainder of 8);

So, the smallest possible value of $$x+y+z$$ is 3+8+9=20.

why z is minimum= 9?
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11 May 2013, 21:27
1
TheNona wrote:
Bunuel wrote:
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

Given $$x=qy+3$$, where $$q$$ is a quotient, an integer $$\geq0$$. Which means that the least value of $$x$$ is when $$q=0$$, in that case $$x=3$$. This basically means that $$x$$ is less than $$y$$. For example 3 divided by 4 yields remainder of 3.

Thus we have that:
$$x$$ is divided by $$y$$ the remainder is 3 --> minimum value of $$x$$ is 3;
$$y$$ is divided by $$z$$ the remainder is 8 --> minimum value of $$y$$ is 8 and minimum value of $$z$$ is one more than 8, so 9 (8 divided by 9 yields the remainder of 8);

So, the smallest possible value of $$x+y+z$$ is 3+8+9=20.

why z is minimum= 9?

When m is divided by n, the remainder can be anything from 0 to (n-1).
Take examples:

When m is divided by 8, the remainder can 0 or 1 or 2 till 7. Can the remainder be 8? No because that means the number is divisible by 8.

Say you divide 20 by 8, you get remainder 4. You divide 21 by 8, you get remainder 5. You divide 22 by 8, you get remainder 6. You divide 23 by 8, you get remainder 7. You divide 24 by 8, you get remainder 0, isn't it? The remainder will not be 8 - it will be 0.

Hence if you know that the remainder when you divide m by n is 7, you can say that n must be at least 8.

Similarly, when a number is divided by z and remainder is 8, z must be at least 9.
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04 Jul 2013, 03:03
Bunuel wrote:
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

$$y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=[color=#ff0000]3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2$$.[/color]

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 $$y$$.

Bunuel, could you please kindly explain this part? How did you come up with this factorization? Many thanks!
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04 Jul 2013, 03:24
BankerRUS wrote:
Bunuel wrote:
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

$$y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=[color=#ff0000]3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2$$.[/color]

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 $$y$$.

Bunuel, could you please kindly explain this part? How did you come up with this factorization? Many thanks!

Consider this: $$(ab-ac)^2=(a(b-c))^2=a^2*(b-c)^2$$, so:

$$(3^5-3^2)^2=(3^2*(3^3-1))^2=(3^2)^2*(3^3-1)^2=3^4*(3^3-1)^2=3^4*26=3^4*2^2*13^2$$.
$$(5^7-5^4)^2=(5^4*(5^3-1))^2=(5^4)^2*(5^3-1)^2=5^8*(5^3-1)^2=5^8*124^2=5^8*2^4*31^2$$

$$(3^4*2^2*13^2)(5^8*2^4*31^2)=2^6*3^4*5^8*13^2*31^2$$.

Hope it's clear.
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18 Sep 2013, 06:00
Bunuel wrote:
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

$$y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2$$.

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 $$y$$.

Can someone please explain how the move from the second part of the equation to the third part was?
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18 Sep 2013, 07:19
ronr34 wrote:
Bunuel wrote:
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

$$y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2$$.

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 $$y$$.

Can someone please explain how the move from the second part of the equation to the third part was?

Consider this: $$(ab-ac)^2=(a(b-c))^2=a^2*(b-c)^2$$, so $$(3^5-3^2)^2=(3^2*(3^3-1))^2=(3^2)^2*(3^3-1)^2=3^4*(3^3-1)^2$$.

Hope it helps.
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08 Oct 2013, 07:32
1
Hi Bunuel,

On Mr. Wallace's Briefcase problem, I was wondering if you can clarify the step where you multiple by 5 choose 3? Can I think of it as SSSNN ["S" representing the number 6 and "N" representing a non-6 digit] and the number of ways to arrange this would be 5!/[3!*2!] since there are 3 repeats for S and 2 repeats for N. Is my reasoning the a logical way to approach this problem as well? Thnks!
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17 Oct 2013, 03:56
nyknicksah20 wrote:
Hi Bunuel,

On Mr. Wallace's Briefcase problem, I was wondering if you can clarify the step where you multiple by 5 choose 3? Can I think of it as SSSNN ["S" representing the number 6 and "N" representing a non-6 digit] and the number of ways to arrange this would be 5!/[3!*2!] since there are 3 repeats for S and 2 repeats for N. Is my reasoning the a logical way to approach this problem as well? Thnks!

Yes, that is correct: $$C^3_5=\frac{5!}{3!2!}$$.
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01 Dec 2013, 03:32
Hello Bunuel,

Since we have all negative numbers in the set, shouldn't we add up the first 5 numbers as the largest?

$$x + (x+2) + (x+4) + (x+6) + (x+8) = 5x +20 = -185$$

$$x = -41$$

Then we take 5 smallest integers, which are the latter in our set

$$(x+4) + (x+6) + (x+8) + (x+10) + (x+12) = 5x +40$$

$$x= - 41$$

$$-41*5 + 40=-165$$

Please correct me if I'm wrong.

Bunuel wrote:
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

Say 7 consecutive odd integers are: $$x$$, $$x+2$$, $$x+4$$, $$x+6$$, $$x+8$$, $$x+10$$, $$x+12$$.

Question: $$x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20=?$$

Given: $$(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=-185$$ --> $$(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=-185$$ --> $$(5x+20)+20=-185$$ --> $$5x+20=-205$$

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01 Dec 2013, 06:54
trailrunner wrote:
Hello Bunuel,

Since we have all negative numbers in the set, shouldn't we add up the first 5 numbers as the largest?

$$x + (x+2) + (x+4) + (x+6) + (x+8) = 5x +20 = -185$$

$$x = -41$$

Then we take 5 smallest integers, which are the latter in our set

$$(x+4) + (x+6) + (x+8) + (x+10) + (x+12) = 5x +40$$

$$x= - 41$$

$$-41*5 + 40=-165$$

Please correct me if I'm wrong.

Bunuel wrote:
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

Say 7 consecutive odd integers are: $$x$$, $$x+2$$, $$x+4$$, $$x+6$$, $$x+8$$, $$x+10$$, $$x+12$$.

Question: $$x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20=?$$

Given: $$(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=-185$$ --> $$(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=-185$$ --> $$(5x+20)+20=-185$$ --> $$5x+20=-205$$

Five largest integers from {$$x$$, $$x+2$$, $$x+4$$, $$x+6$$, $$x+8$$, $$x+10$$, $$x+12$$} are {$$x+4$$, $$x+6$$, $$x+8$$, $$x+10$$, $$x+12$$} no matter whether x is negative or positive.
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14 Feb 2014, 01:39
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, $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$$

Shouldn't -1-|y| be -1-y? since |y| will be positive always?
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17 Feb 2014, 07:50
dnawap123 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, $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$$

Shouldn't -1-|y| be -1-y? since |y| will be positive always?

No.

When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|={-(some \ expression)}$$.

When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|={some \ expression}$$.

Since given that y is a negative number (y<0), then $$|y| = -y$$, hence $$-1-|y|=-1-(-y)=-1+y$$.

Theory on Abolute Values: math-absolute-value-modulus-86462.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.
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03 Jul 2014, 00:52
Bunuel wrote:
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.

Hi Bunnel,

I have a doubt here

I did it in a following way

((8!)^6 ((8!)^4-1))/((8!)^3 ((8!)^2-1)) then I will get (8!)^5 as i can ignore -1. (8!)^4-1 = (8!)^4.

I have seen this kind of solution on some of the problems where you ignore -1 . I did the same here but getting different result.

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03 Jul 2014, 06:09
PathFinder007 wrote:
Bunuel wrote:
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.

Hi Bunnel,

I have a doubt here

I did it in a following way

((8!)^6 ((8!)^4-1))/((8!)^3 ((8!)^2-1)) then I will get (8!)^5 as i can ignore -1. (8!)^4-1 = (8!)^4.

I have seen this kind of solution on some of the problems where you ignore -1 . I did the same here but getting different result.

That's totally wrong. We need to find the tens and units digits of some expression, ignoring anything there will change the result. Where did I say that we can ignore something when we want to find units/tens digit of an expression???
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29 Jul 2014, 05:40
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. $$\frac{yz}{yz+xz-xy}$$

C. $$\frac{yz}{yz+xz+xy}$$

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

E. $$\frac{yz+xz-xy}{yz}$$

With pumps A and B both running and the drain unstopped the pool will be filled in a rate $$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{yz+xz-zy}{xyz}$$ pool/hour. So, the pool will be filled in $$\frac{xyz}{yz+xz-xy}$$ hours (time is reciprocal of rate).

In $$\frac{xyz}{yz+xz-xy}$$ hours A will pump $$\frac{1}{x}*\frac{xyz}{yz+xz-xy}=\frac{yz}{yz+xz-xy}$$ amount of the water into the pool.

can you please tell me if there is any difeerence in these two statements

amount of water in terms of the fraction of the pool which pump A pumped into the pool?

amount of water which pump A pumped into the pool?

and if there is a difference can you please explain?

i had solved the question but was stumped by the language, is it simply asking me the amount of water pumped by A

thanks
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29 Jul 2014, 09:16
tyagigar 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. $$\frac{yz}{yz+xz-xy}$$

C. $$\frac{yz}{yz+xz+xy}$$

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

E. $$\frac{yz+xz-xy}{yz}$$

With pumps A and B both running and the drain unstopped the pool will be filled in a rate $$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{yz+xz-zy}{xyz}$$ pool/hour. So, the pool will be filled in $$\frac{xyz}{yz+xz-xy}$$ hours (time is reciprocal of rate).

In $$\frac{xyz}{yz+xz-xy}$$ hours A will pump $$\frac{1}{x}*\frac{xyz}{yz+xz-xy}=\frac{yz}{yz+xz-xy}$$ amount of the water into the pool.

can you please tell me if there is any difeerence in these two statements

amount of water in terms of the fraction of the pool which pump A pumped into the pool?

amount of water which pump A pumped into the pool?

and if there is a difference can you please explain?

i had solved the question but was stumped by the language, is it simply asking me the amount of water pumped by A

thanks

The first question asks about the fraction of the water pumped by A. For example, if the capacity of the pool is 100 gallons and A pumped 50, then the fraction would be 50/100.

The second question is about the amount. In this case the answer would be 50 gallons.
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07 Sep 2014, 09:30
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.

$$P=\frac{favorable}{total}=\frac{810}{10^5}$$

Hello Bunuel.
My P&C area is kind of weak, and I am working on it. In the solution why is it important to pay attention to no. of ways three 6's can be used in the password? Can you please elaborate on why have you used the Combination formula at all in the solution?

Thanks.
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07 Sep 2014, 09:39
p2bhokie 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.

$$P=\frac{favorable}{total}=\frac{810}{10^5}$$

Hello Bunuel.
My P&C area is kind of weak, and I am working on it. In the solution why is it important to pay attention to no. of ways three 6's can be used in the password? Can you please elaborate on why have you used the Combination formula at all in the solution?

Thanks.

We have 5-digit password: ABCDE, out of which there are three 6's. These 6's can take any place: ABC, ABD, ABE, ... $$C^3_5$$ gives the number of ways to choose which 3 digits (which 3 letters out of 5) will be 6's:
666DE
66C6E
...
_________________
Re: Baker's Dozen &nbs [#permalink] 07 Sep 2014, 09:39

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