In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
In how many ways can a person post 5 letters in 3 letter box
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IMPORTANT: The question doesn't specify whether the letters and the letter boxes are UNIQUE, but I'm going to assume that they are unique.mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
Take the task of distributing the 5 letters and break it into stages.
Stage 1: Select a box for the 1st letter to go into.
There are 3 available boxes, so we can complete stage 1 in 3 ways
Stage 2: Select a box for the 2nd letter to go into.
There are 3 available boxes, so we can complete stage 2 in 3 ways
Stage 3: Select a box for the 3rd letter to go into.
There are 3 available boxes, so we can complete stage 3 in 3 ways
Stage 4: Select a box for the 4th letter to go into.
There are 3 available boxes, so we can complete stage 4 in 3 ways
Stage 5: Select a box for the 5th letter to go into.
There are 3 available boxes, so we can complete stage 5 in 3 ways
By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus distribute all 5 letters) in (3)(3)(3)(3)(3) ways (= 3â�µ ways)
Answer: D

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Cheers,
Brent

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Apparently, the solution is quite simple here:mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
each letter will have 3 boxes to choose from and so answer 3^5.
But I am having a tough time visualizing this solution. The way I am looking at is:
1box 2box 3box
5 0 0
4 0 1
3 0 2
3 1 1
2 2 1
Now these can be arranged internally for example  500 could be 005 etc.
What mistake am I making? Thanks
GMAT/MBA Expert
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If that's your final solution, then you appear to be treating the 5 letters as 5 IDENTICAL letters, and you are treating the 3 mailboxes as IDENTICAL mailboxesmensanumber wrote:Apparently, the solution is quite simple here:mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
each letter will have 3 boxes to choose from and so answer 3^5.
But I am having a tough time visualizing this solution. The way I am looking at is:
1box 2box 3box
5 0 0
4 0 1
3 0 2
3 1 1
2 2 1
Now these can be arranged internally for example  500 could be 005 etc.
What mistake am I making? Thanks
Cheers,
Brent
Not making any mistakes so far but haven't completed the thought.mensanumber wrote:Apparently, the solution is quite simple here:mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
each letter will have 3 boxes to choose from and so answer 3^5.
But I am having a tough time visualizing this solution. The way I am looking at is:
1box 2box 3box
5 0 0
4 0 1
3 0 2
3 1 1
2 2 1
Now these can be arranged internally for example  500 could be 005 etc.
What mistake am I making? Thanks
As you suggest, 500 can be arranged 3 ways.
Likewise, 401 can be arranged 3 ways.
And so on. All together there are then 3x3x3x3x3 = 3^5 ways

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Hi Brent,[email protected] wrote:If that's your final solution, then you appear to be treating the 5 letters as 5 IDENTICAL letters, and you are treating the 3 mailboxes as IDENTICAL mailboxesmensanumber wrote:Apparently, the solution is quite simple here:mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
each letter will have 3 boxes to choose from and so answer 3^5.
But I am having a tough time visualizing this solution. The way I am looking at is:
1box 2box 3box
5 0 0
4 0 1
3 0 2
3 1 1
2 2 1
Now these can be arranged internally for example  500 could be 005 etc.
What mistake am I making? Thanks
Cheers,
Brent
Thanks for your reply.
That's not my full solutions. Here it is:
1box 2box 3box
5 0 0......................................Case1
4 0 1......................................Case2
3 0 2......................................Case3
3 1 1......................................Case4
2 2 1......................................Case5
Let's consider each case individually,
Case1
500 i.e. 5 letters in box1, 0 letters in box2, 0 letters box3. Now, there will be 3 such scenarios, 3P1/2P1 = 3. (by MISSISSIPPI rule dividing by 2P1 for two identical zeros) These 3 scenarios are 500, 050, 005
Since letters distinct, for each of these scenarios: 5C5*0C0*0C0
So, total possible arrangements for case1 : (3P1/2P1)*5C5*0C0*0C0 = 3*1*1*1 = 3
Similarly,
Case2: 401, total possible arrangements : (3P1)*5C4*1C0*1C1 = 6*5*1*1 = 30
Case3: 302, total possible arrangements : (3P1)*5C3*2C0*2C2 = 6*10*1*1 = 60
Case4: 311, total possible arrangements : (3P1/2P1)*5C3*2C1*1C1 = 3*10*2*1 = 60
Case4: 221, total possible arrangements : (3P1/2P1)*5C2*3C2*1C1 = 3*10*3*1 = 90
Total of all cases = 3+30+60+60+90 = 243 which is indeed 3^5
But what irritates me is my long winded solution is visualizeable to me and not your more elegant solution.
Can you help me see it from your point of view?
Thanks again

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Thanks for your reply Regor.regor60 wrote:Not making any mistakes so far but haven't completed the thought.mensanumber wrote:Apparently, the solution is quite simple here:mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
each letter will have 3 boxes to choose from and so answer 3^5.
But I am having a tough time visualizing this solution. The way I am looking at is:
1box 2box 3box
5 0 0
4 0 1
3 0 2
3 1 1
2 2 1
Now these can be arranged internally for example  500 could be 005 etc.
What mistake am I making? Thanks
As you suggest, 500 can be arranged 3 ways.
Likewise, 401 can be arranged 3 ways.
And so on. All together there are then 3x3x3x3x3 = 3^5 ways
Here is my full solution:
1box 2box 3box
5 0 0......................................Case1
4 0 1......................................Case2
3 0 2......................................Case3
3 1 1......................................Case4
2 2 1......................................Case5
Let's consider each case individually,
Case1
500 i.e. 5 letters in box1, 0 letters in box2, 0 letters box3. Now, there will be 3 such scenarios, 3P1/2P1 = 3. (by MISSISSIPPI rule dividing by 2P1 for two identical zeros) These 3 scenarios are 500, 050, 005
Since letters distinct, for each of these scenarios: 5C5*0C0*0C0
So, total possible arrangements for case1 : (3P1/2P1)*5C5*0C0*0C0 = 3*1*1*1 = 3
Similarly,
Case2: 401, total possible arrangements : (3P1)*5C4*1C0*1C1 = 6*5*1*1 = 30
Case3: 302, total possible arrangements : (3P1)*5C3*2C0*2C2 = 6*10*1*1 = 60
Case4: 311, total possible arrangements : (3P1/2P1)*5C3*2C1*1C1 = 3*10*2*1 = 60
Case4: 221, total possible arrangements : (3P1/2P1)*5C2*3C2*1C1 = 3*10*3*1 = 90
Total of all cases = 3+30+60+60+90 = 243 which is indeed 3^5
But what irritates me is my long winded solution is visualizeable to me and not your more elegant solution.
Can you help me see it from your point of view?
Thanks again
I read your response too quickly so ignore what I wrote, it's not correct. The way you've done it is correct.mensanumber wrote:Thanks for your reply Regor.regor60 wrote:Not making any mistakes so far but haven't completed the thought.mensanumber wrote:Apparently, the solution is quite simple here:mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
each letter will have 3 boxes to choose from and so answer 3^5.
But I am having a tough time visualizing this solution. The way I am looking at is:
1box 2box 3box
5 0 0
4 0 1
3 0 2
3 1 1
2 2 1
Now these can be arranged internally for example  500 could be 005 etc.
What mistake am I making? Thanks
As you suggest, 500 can be arranged 3 ways.
Likewise, 401 can be arranged 3 ways.
And so on. All together there are then 3x3x3x3x3 = 3^5 ways
Here is my full solution:
1box 2box 3box
5 0 0......................................Case1
4 0 1......................................Case2
3 0 2......................................Case3
3 1 1......................................Case4
2 2 1......................................Case5
Let's consider each case individually,
Case1
500 i.e. 5 letters in box1, 0 letters in box2, 0 letters box3. Now, there will be 3 such scenarios, 3P1/2P1 = 3. (by MISSISSIPPI rule dividing by 2P1 for two identical zeros) These 3 scenarios are 500, 050, 005
Since letters distinct, for each of these scenarios: 5C5*0C0*0C0
So, total possible arrangements for case1 : (3P1/2P1)*5C5*0C0*0C0 = 3*1*1*1 = 3
Similarly,
Case2: 401, total possible arrangements : (3P1)*5C4*1C0*1C1 = 6*5*1*1 = 30
Case3: 302, total possible arrangements : (3P1)*5C3*2C0*2C2 = 6*10*1*1 = 60
Case4: 311, total possible arrangements : (3P1/2P1)*5C3*2C1*1C1 = 3*10*2*1 = 60
Case4: 221, total possible arrangements : (3P1/2P1)*5C2*3C2*1C1 = 3*10*3*1 = 90
Total of all cases = 3+30+60+60+90 = 243 which is indeed 3^5
But what irritates me is my long winded solution is visualizeable to me and not your more elegant solution.
Can you help me see it from your point of view?
Thanks again
GMAT/MBA Expert
 [email protected]
 GMAT Instructor
 Posts: 1460
 Joined: 09 Apr 2015
 Location: New York, NY
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Since each letter can be put into any of the 3 boxes, then each letter has 3 choices. Thus, the number of ways a person can put 5 letters in 3 boxes is 3 x 3 x 3 x 3 x 3 = 3^5.mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
Answer: D
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