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letters can be posted

by sanju09 » Mon Apr 04, 2011 2:35 am
In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?
(A) 5C3
(B) 2^5
(C) 5P3
(D) 5^3
(E) 3^5
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by manpsingh87 » Mon Apr 04, 2011 2:59 am
sanju09 wrote:In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?
(A) 5C3
(B) 2^5
(C) 5P3
(D) 5^3
(E) 3^5
1st letter can go to any of the 3 boxes in 3 ways, similarly 2nd letter can go in any of 3 post boxes in 3 ways, also for remaining, hence total no. of ways = 3*3*3*3*3=3^5 hence E
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by force5 » Mon Apr 04, 2011 4:44 am
3^5 E
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by preeti6 » Mon Apr 04, 2011 8:14 pm
Can you please clarify as to why the answer cannot be D? There are three boxes and 5 letters can be placed in any arrangement amongst them.. so the first box can have 5 letters, same with the second and thrid box, so 5^3?
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by Anurag@Gurome » Mon Apr 04, 2011 8:21 pm
preeti6 wrote:Can you please clarify as to why the answer cannot be D? There are three boxes and 5 letters can be placed in any arrangement amongst them.. so the first box can have 5 letters, same with the second and thrid box, so 5^3?
What you are doing implies that 1 letter is going to more than 1 box, which is not possible.
Note that 1 box can hold more than 1 letter but 1 letter cannot go to more than 1 box.
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by manpsingh87 » Mon Apr 04, 2011 8:25 pm
preeti6 wrote:Can you please clarify as to why the answer cannot be D? There are three boxes and 5 letters can be placed in any arrangement amongst them.. so the first box can have 5 letters, same with the second and thrid box, so 5^3?
as per the question
any number of letters can be posted in all of the three post boxes
we have to post the letter in to the boxes therefore we must make arrangements from the letters perspective..!!!

i hope it helps..!!!!
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by preeti6 » Mon Apr 04, 2011 9:02 pm
@all - thanks! that was helpful.. it makes sense to me nw!
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by muthu8622 » Thu May 12, 2011 1:52 am
what if take the arrangement like this..

(1st box, 2nd box, 3rd box)

(5,0,0) (4,1,0) (3,2,0) (2,3,0) (1,4,0) (0,5,0)
(4,0,1) (3,1,1) (2,2,1) (1,3,1) (0,4,1)
(3,0,2) (2,1,2) (1,2,2) (0,3,2)
(2,0,3) (1,1,3) (0,2,3)
(1,0,4) (0,1,4)
(0,0,5)
1+2+3+4+5+6 = 21 ways

Whats wrong with this answer? i'm confused why this is not accepeted.. need help thanks
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by Brent@GMATPrepNow » Thu May 12, 2011 6:19 am
muthu8622 wrote:what if take the arrangement like this..

(1st box, 2nd box, 3rd box)

(5,0,0) (4,1,0) (3,2,0) (2,3,0) (1,4,0) (0,5,0)
(4,0,1) (3,1,1) (2,2,1) (1,3,1) (0,4,1)
(3,0,2) (2,1,2) (1,2,2) (0,3,2)
(2,0,3) (1,1,3) (0,2,3)
(1,0,4) (0,1,4)
(0,0,5)
1+2+3+4+5+6 = 21 ways

Whats wrong with this answer? i'm confused why this is not accepeted.. need help thanks
Nice work, muthu8622

I like how you've set up this solution - very systematic

Aside: students often reject counting solutions where the outcomes are listed, but this strategy often helps us see a pattern in the solution (as there is here)

The problem isn't with your solution; the problem is with the question.

In your solution, you're treating the letters as identical, and other posters here are treating each letter as unique.

If the letters are identical (say junk mail) then the number of outcomes is 21

If each letter is unique (different from the others) then the number of outcomes is 3^5

Given the ambiguous nature of the question, it's hard to say what the "official" answer is. Also, given its ambiguity, this could never be an official GMAT question.

Cheers,
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by Brent@GMATPrepNow » Thu May 12, 2011 6:27 am
sanju09 wrote:In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?
(A) 5C3
(B) 2^5
(C) 5P3
(D) 5^3
(E) 3^5
I should also mention that the GMAT does not require students to know the notation for combinations (nCr) and permutations (nPr).

The main reason is that it's often possible to solve counting questions using techniques other than applying the rules for combinations and permutations. So, giving answers with this notation would be too restrictive.

Another reason is that counting notation is not universal. For example, while some people might use the notation 5C2 to represent the idea of "5 choose 2," others use bracket notation as shown below.

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by djiddish98 » Thu May 12, 2011 7:44 am
Thanks for the clarification, Brent. I used the dividers method for this and got 21 as well.

|| LLLLL

7!/(2!5!)
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by Brent@GMATPrepNow » Thu May 12, 2011 9:01 am
djiddish98 wrote:Thanks for the clarification, Brent. I used the dividers method for this and got 21 as well.

|| LLLLL

7!/(2!5!)
Nice work - the dividers method is a very advanced approach!

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by muthu8622 » Fri May 13, 2011 2:30 am
Brent@GMATPrepNow wrote:
muthu8622 wrote:what if take the arrangement like this..

(1st box, 2nd box, 3rd box)

(5,0,0) (4,1,0) (3,2,0) (2,3,0) (1,4,0) (0,5,0)
(4,0,1) (3,1,1) (2,2,1) (1,3,1) (0,4,1)
(3,0,2) (2,1,2) (1,2,2) (0,3,2)
(2,0,3) (1,1,3) (0,2,3)
(1,0,4) (0,1,4)
(0,0,5)
1+2+3+4+5+6 = 21 ways

Whats wrong with this answer? i'm confused why this is not accepeted.. need help thanks
Nice work, muthu8622

I like how you've set up this solution - very systematic

Aside: students often reject counting solutions where the outcomes are listed, but this strategy often helps us see a pattern in the solution (as there is here)

The problem isn't with your solution; the problem is with the question.

In your solution, you're treating the letters as identical, and other posters here are treating each letter as unique.

If the letters are identical (say junk mail) then the number of outcomes is 21

If each letter is unique (different from the others) then the number of outcomes is 3^5

Given the ambiguous nature of the question, it's hard to say what the "official" answer is. Also, given its ambiguity, this could never be an official GMAT question.

Cheers,
Brent

Thanks Brent for clarification of the question, i was wondering where i went wrong, now i know it's depending on how you interpret

one more thing how do you use the dividers method
7!/(2!5!) = 21
i would like to know why its 7! when there is only 5 letters.. a bit confused here.. i've been trying to figure it for quite some time to find easy way other than counting method..

Especially on questions like 10 letters to be posted on 5 post boxes, the pattern would be
1
1+3
1+3+6
1+3+6+10
1+3+6+10+15
1+3+6+10+15+21
1+3+6+10+15+21+28
1+3+6+10+15+21+28+36
1+3+6+10+15+21+28+36+45
1+3+6+10+15+21+28+36+45+55
1+3+6+10+15+21+28+36+45+55+66
Total = 1001
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by djiddish98 » Fri May 13, 2011 4:39 am
To my knowledge the dividers method works by taking accounting for all the ways you can arrange X objects into Y bins (assuming the order doesn't matter)

we have 5 letters, which we can denote as L L L L L

We also have 3 mailboxes. To graphically represent these mailboxes, we create the dividers, which I'll denote as |

In order to break up the objects (5 L's) in 3 mail boxes, we just need 2 dividers. The dividers partition the boxes; they are not the boxes themselves.

Notice that L|LL|LL takes the 5 letters and put them into 3 separate boxes - 1 in the first box, 2 in the second box, 2 in the third box.

We can also have ||LLLLL, which puts all 5 letters into the 3rd mail box and leaves the other two empty.

So we want to know how many ways we can arrange the two dividers.

If we add the count of the dividers and the count of the letters, we get 7. And we "choose" 2 dividers, so the function is 7 C 2 or 7!/(2!(7-2)!)

I believe the universal way to view it in this example would be (Letters + Boxes -1) C (Boxes - 1)

10 letters to 5 boxes would be 14 c 4 then, which coincides with your answer of 1001.

14!/(10!)(4!) -> 14*13*12*11 / (12*2*1) -> 7*13*11
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by muthu8622 » Fri May 13, 2011 5:52 am
djiddish98 wrote:To my knowledge the dividers method works by taking accounting for all the ways you can arrange X objects into Y bins (assuming the order doesn't matter)

we have 5 letters, which we can denote as L L L L L

We also have 3 mailboxes. To graphically represent these mailboxes, we create the dividers, which I'll denote as |

In order to break up the objects (5 L's) in 3 mail boxes, we just need 2 dividers. The dividers partition the boxes; they are not the boxes themselves.

Notice that L|LL|LL takes the 5 letters and put them into 3 separate boxes - 1 in the first box, 2 in the second box, 2 in the third box.

We can also have ||LLLLL, which puts all 5 letters into the 3rd mail box and leaves the other two empty.

So we want to know how many ways we can arrange the two dividers.

If we add the count of the dividers and the count of the letters, we get 7. And we "choose" 2 dividers, so the function is 7 C 2 or 7!/(2!(7-2)!)

I believe the universal way to view it in this example would be (Letters + Boxes -1) C (Boxes - 1)

10 letters to 5 boxes would be 14 c 4 then, which coincides with your answer of 1001.

14!/(10!)(4!) -> 14*13*12*11 / (12*2*1) -> 7*13*11

Thanks a lot djiddish98Really Appreciate your explanation.. now i understand the use of dividers method..
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