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# Baker's Dozen

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Baker's Dozen [#permalink]  08 Mar 2012, 12:27
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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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Re: Baker's Dozen [#permalink]  14 Mar 2012, 16:10
Expert's post
onedayill 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

Hi Bunuel,

I didn't quite understand this ..

X and Y are odd then how can |x| = -x it has to be +ive.

x and y are negative numbers, not odd numbers.

If x\geq{0} then |x|=x;
If x<{0} then |x|=-x. So if x is negative then |x|=-x=-negative=positive. For example, if x=-2 then |x|=|-2|=2=-x.

Check Absolute Values chapter of Math Book for more: math-absolute-value-modulus-86462.html

Hope it helps.
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Re: Baker's Dozen [#permalink]  14 Mar 2012, 22:43
Got 9 correct with @ 2.5 mins/Q ... how about others ? (2 Q left without going to deep as killed more than 4 mins on those Q4,7- this proves that not to waste time on something you don't get idea in 1.5-2 mins time frame and missed 2 narrowly- Q9, 10: 10 I din't read 2y but only y so...)..will post in details some explanations, esp. those which are little different from already posted by the very own legend Mr B!!
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Re: Baker's Dozen [#permalink]  16 Mar 2012, 02:47
Expert's post
chetan2u 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 fraction of the pool which pump A filled?
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 pump A will do \frac{1}{x}*\frac{xyz}{yz+xz-xy}=\frac{yz}{yz+xz-xy} part of the job.

hi , i feel the ans to this Q is wrong. as we have to tell what fraction of the pool was filled up by pump A. it will not depend on the drain C as the filling was done only by pump A and B. ans shud be y/(x+y).
the given ans would be correct if we were to find "the amount of water pumped by A in that duration"

Actually that was the intended meaning of the question. I edited it so to avoid ambiguity: "which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool?"

Hope now it's more precise.
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Re: Baker's Dozen [#permalink]  16 Mar 2012, 06:01
Hi Bunuel,
Please help me in understanding the below text. I think I am lacking this concept.I always try to solve by taking LCM in order to get rid of the fractions. However, I found your approach short n simple. If you could shed some light on this, would be really helpful

Fritz owns \frac{2}{2+3}=\frac{2}{5}th of all shares

Thanks
H

Bunuel wrote:
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

Fritz owns is \frac{2}{3}rd of the shares of the other three shareholders --> Fritz owns \frac{2}{2+3}=\frac{2}{5}th of all shares;
Luis owns is \frac{3}{7}th of the shares of the other three shareholders --> Luis owns \frac{3}{3+7}=\frac{3}{10}th of all shares;
Alfred owns is \frac{4}{11}th of the shares of the other three shareholders --> Alfred owns \frac{4}{4+11}=\frac{4}{15}th of all shares;

Together those three own \frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}th of all shares, which means that Werner owns 1-\frac{29}{30}=\frac{1}{30}. Hence from $3,600,000 Werner gets$3,600,000*\frac{1}{30}=$120,000. Answer: D. _________________ +1 Kudos me, Help me unlocking GMAT Club Tests Math Expert Joined: 02 Sep 2009 Posts: 18741 Followers: 3250 Kudos [?]: 22427 [0], given: 2619 Re: Baker's Dozen [#permalink] 17 Mar 2012, 00:12 Expert's post GMATD11 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=2^4*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. Pls correct the Typo 2^4 to 2^8 Typo corrected, thanks. _________________ Intern Joined: 09 Feb 2012 Posts: 11 Followers: 0 Kudos [?]: 0 [0], given: 21 Re: Baker's Dozen [#permalink] 18 Mar 2012, 15:45 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. Hello Bunnel.... Can you kindly clarify this question and elaborate on the solution!! Thanks in advance. Math Expert Joined: 02 Sep 2009 Posts: 18741 Followers: 3250 Kudos [?]: 22427 [0], given: 2619 Re: Baker's Dozen [#permalink] 19 Mar 2012, 03:10 Expert's post balas 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. Hello Bunnel.... Can you kindly clarify this question and elaborate on the solution!! Thanks in advance. Can you please specify which part didn't you understand? Thanks. _________________ Manager Joined: 22 Feb 2012 Posts: 93 Schools: HBS '16 GMAT 1: 740 Q49 V42 GMAT 2: 670 Q42 V40 GPA: 3.47 WE: Corporate Finance (Aerospace and Defense) Followers: 2 Kudos [?]: 19 [0], given: 25 Re: Baker's Dozen [#permalink] 27 Mar 2012, 10:04 Awesome set of questions! Intern Joined: 14 Apr 2012 Posts: 7 Followers: 0 Kudos [?]: 0 [0], given: 35 Re: Baker's Dozen [#permalink] 28 Apr 2012, 02:41 Can any one please explain: why it is (9*9*1*1*1* 5C3) in Question no. 1 (password problem)? Why not : 9*9*1*1*1* 5!/2! ?? Last edited by mofasser08 on 28 Apr 2012, 03:00, edited 1 time in total. Manager Joined: 12 Feb 2012 Posts: 106 Followers: 1 Kudos [?]: 10 [0], given: 28 Re: Baker's Dozen [#permalink] 28 Apr 2012, 18:47 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. Hey Bunuel, Sorry to trouble you. If you could, I would appreciate if you could clarify more about 5C3. So this is how I approached the problem: We have three - 6's. With two slots that can be filled by 9 numbers. 666 9 9 . Now the question is how many ways can we rearrange these codes of three 6's and 2 non 6's? Well, have 5 elements 3 identical (because we have 3 identical 6's and two non-6's) Hence we have 5!/(3!2!) ways of rearranging the numbers. Is this why we have 5C3 in the problem? But the two non-6's can be identical or different. For example 66611 66622 66633 66644 66655 ... 66699... etc all satisfy the condition of only three 6's. The number of ways to rearrange the letters {aaabb} is 5!/(3!2!). Moreover, we can also have the two non-6's be different, 66612 66654 66624, etc . the number of ways to rearrange the letters {aaabc} is 5!/(3!). What do we do in this situation? Do we look at this problem as (#6)(#6)(#6)(not #6)(not #6) therefore 5!/(3!2!). Three repeating elements of #6 and two repeating elements of not #6. OR Do we look at the problem as (#6)(#6)(#6)(x)(y), where x and y two numbers that are not 6, that may or may not be equal, therefore 5!/3! (3! for the 3 repeating 6's but we cannot say anything about whether x and y are equal to each other) Bunuel, as always, thank you so much!! And again, sorry to trouble you. Manager Joined: 12 Feb 2012 Posts: 106 Followers: 1 Kudos [?]: 10 [0], given: 28 Re: Baker's Dozen [#permalink] 28 Apr 2012, 18:51 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*1*1*1* 5!/2! ?? Mofasser, We pretty much have the same question. But I think the reasons it's 5C3 is because we have (#6)(#6)(#6)(not #6)(not #6)=aaabb. The number of ways to rearrange aaabb is 5!/(3!2!). But we need to check with the Oracle, Bunuel. Intern Joined: 01 Mar 2012 Posts: 32 Concentration: Operations, Finance GMAT 1: 740 Q49 V41 GPA: 3.3 WE: Engineering (Manufacturing) Followers: 1 Kudos [?]: 3 [0], given: 46 Re: Baker's Dozen [#permalink] 02 May 2012, 09:59 Question 1 3 number of 6's can be arranged among themselves in 1 way. Number of ways we can select three position from 5= 5C3. For the remaining two position we can choose any digit but 6. So number of possible patterns=(5C3)*9*9 Hence required probability=(5C3)*9*9/10^5 I hope those who have doubt can understand this explanation Senior Manager Joined: 21 Mar 2010 Posts: 316 Followers: 5 Kudos [?]: 20 [0], given: 33 Re: Baker's Dozen [#permalink] 08 May 2012, 17:46 Bunuel wrote: 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 Fritz owns is \frac{2}{3}rd of the shares of the other three shareholders --> Fritz owns \frac{2}{2+3}=\frac{2}{5}th of all shares; Luis owns is \frac{3}{7}th of the shares of the other three shareholders --> Luis owns \frac{3}{3+7}=\frac{3}{10}th of all shares; Alfred owns is \frac{4}{11}th of the shares of the other three shareholders --> Alfred owns \frac{4}{4+11}=\frac{4}{15}th of all shares; Together those three own \frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}th of all shares, which means that Werner owns 1-\frac{29}{30}=\frac{1}{30}. Hence from$3,600,000 Werner gets $3,600,000*\frac{1}{30}=$120,000.

Could some one explain this to me how does one arrive at 2/2+3 = 2/5 and so on?
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Re: Baker's Dozen [#permalink]  09 May 2012, 00:10
Expert's post
mbafall2011 wrote:
Bunuel wrote:
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

Fritz owns is \frac{2}{3}rd of the shares of the other three shareholders --> Fritz owns \frac{2}{2+3}=\frac{2}{5}th of all shares;
Luis owns is \frac{3}{7}th of the shares of the other three shareholders --> Luis owns \frac{3}{3+7}=\frac{3}{10}th of all shares;
Alfred owns is \frac{4}{11}th of the shares of the other three shareholders --> Alfred owns \frac{4}{4+11}=\frac{4}{15}th of all shares;

Together those three own \frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}th of all shares, which means that Werner owns 1-\frac{29}{30}=\frac{1}{30}. Hence from $3,600,000 Werner gets$3,600,000*\frac{1}{30}=\$120,000.

Could some one explain this to me how does one arrive at 2/2+3 = 2/5 and so on?

Explained here: baker-s-dozen-128782-40.html#p1059585
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Re: Baker's Dozen [#permalink]  04 Jun 2012, 10:16
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.

Hi Bunuel..can you tell me if I am thinking correct.

1/10 * 1/10 * 1/10 * 9/10 * 9/10 this gives me 3 chances of getting 6 and rest 2 for getting any other number.
now the above scenario can be arranged in 5C3 ways so

1/10 * 1/10 * 1/10 * 9/10 * 9/10 * 5C3..

I hope I am correct in this way..?

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Re: Baker's Dozen [#permalink]  12 Jun 2012, 17:39
Hi,
Thanks!
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Re: Baker's Dozen [#permalink]  13 Jun 2012, 01:52
Expert's post
sharmila79 wrote:
Hi,
Thanks!

Switch view mode of the topic from "Best Reply" to "Oldest" and the links from the initial post will lead you to the posts with solutions. The question about the shareholders is answered here: baker-s-dozen-128782-40.html#p1059585

Hope it helps.
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Re: Baker's Dozen [#permalink]  13 Jun 2012, 05:11
Thank you very much! I got it.
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Re: Baker's Dozen [#permalink]  16 Jun 2012, 05:11
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}

Hi Bunnel,

I didn't understand how you got 5C3, can you please explain more this...
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Re: Baker's Dozen [#permalink]  16 Jun 2012, 05:15
Expert's post
kotela 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}

Hi Bunnel,

I didn't understand how you got 5C3, can you please explain more this...

Check this: baker-s-dozen-128782-60.html#p1079496

Hope it helps.
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Re: Baker's Dozen   [#permalink] 16 Jun 2012, 05:15
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