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Baker's Dozen

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Baker's Dozen  [#permalink]

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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 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.

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Re: Baker's Dozen  [#permalink]

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New post 27 Mar 2017, 18:07
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Here's my solution for #2 (see visual below). It helps to use brackets!
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Baker's Dozen  [#permalink]

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New post 19 Dec 2017, 03:51
Bunuel niks18

Quote:
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



Quote:
Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\).


I solved the same problem taking hint from number theory that a consecutive odd number is represented by:

2x+1 , 2x+3

My approach had slightly complex calculations but it ensured that my first no in series is always odd.

Is this understanding correct? Or can I still take a random integer x as start and then add +2 (to get consecutive ones)
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Re: Baker's Dozen  [#permalink]

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New post 19 Dec 2017, 04:03
adkikani wrote:
Bunuel niks18

Quote:
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



Quote:
Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\).


I solved the same problem taking hint from number theory that a consecutive odd number is represented by:

2x+1 , 2x+3

My approach had slightly complex calculations but it ensured that my first no in series is always odd.

Is this understanding correct? Or can I still take a random integer x as start and then add +2 (to get consecutive ones)


In this question, as shown, you can take 7 consecutive odd integers to be \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), and \(x+12\). For some number properties questions, you should represent odd integers, as 2k + 1, 2k + 3, ... Here you can also take the integers, to be 2k + 1, 2k + 3, ... but it's not necessary.
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Re: Baker's Dozen  [#permalink]

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New post 20 Mar 2019, 07:13
C812−(C77∗C15+C55∗C37)=495−(5+35)=455C128−(C77∗C51+C55∗C73)=495−(5+35)=455.

How do you get (5+35) (I know that it's probably very simple...)? Bunuel

Thank you in advance!!
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Re: Baker's Dozen  [#permalink]

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New post 25 Aug 2019, 02:28
Bunuel wrote:
mofasser08 wrote:
Can any one please explain: why it is (9*9*1*1*1* 5C3) in Question no. 1 (password problem)? Why not : 9*9*5P3 ??


That's a tough 700+ problem.

We are interested 666XX numbers. Now, two this can be arranged in 10 ways;
666XX;
66X6X;
6X66X;
X666X;

66XX6;
6X6X6;
X66X6;

6XX66;
XX666;

X6X66.

So, basically with \(C^3_5=10\) we are choosing which 3 places out of 5 will be occupied by X's. Or we could do the other way around: \(C^2_510\) choosing which 2 places out of 5 will be occupied by 6's.

Next, 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 5-digit password unlike 5-digit number can start with 0.

Total \(9*9*C^3_5=810\).

Hope it's clear.




Bunuel

1.For XXXYY arrangement – 1*1*1*9*1*5!/(3!2!) = 90
2.For XXXYZ arrangement – 1*1*1*9*8*5!/3! = 1440
Hence total arrangement 90+1440=1530
Please explain where I am making a mistake

I think if we have three 6 and two 9 then we get

666XX;
66X6X;
6X66X;
X666X;

66XX6;
6X6X6;
X66X6;

6XX66;
XX666;

X6X66.

10 cases

But if we have three 6 and two different numbers such as 1,2 then we get 20 cases .
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Re: Baker's Dozen   [#permalink] 25 Aug 2019, 02:28

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