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# In how many ways can a person post 5 letters in 3 letter box

tagged by: mensanumber

#### In how many ways can a person post 5 letters in 3 letter box

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

### GMAT/MBA Expert

GMAT Instructor
Joined
08 Dec 2008
Posted:
12122 messages
Followed by:
1237 members
5254
GMAT Score:
770
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
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.

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 (= 3⁵ ways)

--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting?id=775

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- http://www.beatthegmat.com/permutation-combination-simple-one-t257412.html
- http://www.beatthegmat.com/simple-one-t270061.html
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MEDIUM
- http://www.beatthegmat.com/combinatorics-solution-explanation-t273194.html
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- http://www.beatthegmat.com/sub-sets-probability-t273337.html
- http://www.beatthegmat.com/combinatorics-problem-t273180.html
- http://www.beatthegmat.com/digits-numbers-t270127.html
- http://www.beatthegmat.com/doubt-on-separator-method-t271047.html
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DIFFICULT
- http://www.beatthegmat.com/wonderful-p-c-ques-t271001.html
- http://www.beatthegmat.com/ps-counting-t273659.html
- http://www.beatthegmat.com/permutation-and-combination-t273915.html

Cheers,
Brent

_________________
Brent Hanneson – Creator of GMATPrepNow.com
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Senior | Next Rank: 100 Posts
Joined
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Posted:
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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
Apparently, the solution is quite simple here:
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

GMAT Instructor
Joined
08 Dec 2008
Posted:
12122 messages
Followed by:
1237 members
5254
GMAT Score:
770
mensanumber wrote:
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
Apparently, the solution is quite simple here:
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
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 mailboxes

Cheers,
Brent

_________________
Brent Hanneson – Creator of GMATPrepNow.com
Use our video course along with

And check out all of our free resources

GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!

### Top Member

Master | Next Rank: 500 Posts
Joined
15 Oct 2009
Posted:
308 messages
27
mensanumber wrote:
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
Apparently, the solution is quite simple here:
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
Not making any mistakes so far but haven't completed the thought.

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

Senior | Next Rank: 100 Posts
Joined
23 Oct 2013
Posted:
30 messages
Followed by:
1 members
2
GMAT Score:
750
Brent@GMATPrepNow wrote:
mensanumber wrote:
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
Apparently, the solution is quite simple here:
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
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 mailboxes

Cheers,
Brent
Hi Brent,

That's not my full solutions. Here it is:

1box 2box 3box
5 0 0......................................Case-1
4 0 1......................................Case-2
3 0 2......................................Case-3
3 1 1......................................Case-4
2 2 1......................................Case-5

Let's consider each case individually,

Case-1
500 i.e. 5 letters in box1, 0 letters in box-2, 0 letters box-3. 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 case-1 : (3P1/2P1)*5C5*0C0*0C0 = 3*1*1*1 = 3

Similarly,
Case-2: 401, total possible arrangements : (3P1)*5C4*1C0*1C1 = 6*5*1*1 = 30
Case-3: 302, total possible arrangements : (3P1)*5C3*2C0*2C2 = 6*10*1*1 = 60
Case-4: 311, total possible arrangements : (3P1/2P1)*5C3*2C1*1C1 = 3*10*2*1 = 60
Case-4: 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 visualize-able to me and not your more elegant solution.
Can you help me see it from your point of view?

Thanks again

Senior | Next Rank: 100 Posts
Joined
23 Oct 2013
Posted:
30 messages
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1 members
2
GMAT Score:
750
regor60 wrote:
mensanumber wrote:
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
Apparently, the solution is quite simple here:
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
Not making any mistakes so far but haven't completed the thought.

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......................................Case-1
4 0 1......................................Case-2
3 0 2......................................Case-3
3 1 1......................................Case-4
2 2 1......................................Case-5

Let's consider each case individually,

Case-1
500 i.e. 5 letters in box1, 0 letters in box-2, 0 letters box-3. 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 case-1 : (3P1/2P1)*5C5*0C0*0C0 = 3*1*1*1 = 3

Similarly,
Case-2: 401, total possible arrangements : (3P1)*5C4*1C0*1C1 = 6*5*1*1 = 30
Case-3: 302, total possible arrangements : (3P1)*5C3*2C0*2C2 = 6*10*1*1 = 60
Case-4: 311, total possible arrangements : (3P1/2P1)*5C3*2C1*1C1 = 3*10*2*1 = 60
Case-4: 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 visualize-able to me and not your more elegant solution.
Can you help me see it from your point of view?

Thanks again

### Top Member

Master | Next Rank: 500 Posts
Joined
15 Oct 2009
Posted:
308 messages
27
mensanumber wrote:
regor60 wrote:
mensanumber wrote:
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
Apparently, the solution is quite simple here:
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
Not making any mistakes so far but haven't completed the thought.

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......................................Case-1
4 0 1......................................Case-2
3 0 2......................................Case-3
3 1 1......................................Case-4
2 2 1......................................Case-5

Let's consider each case individually,

Case-1
500 i.e. 5 letters in box1, 0 letters in box-2, 0 letters box-3. 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 case-1 : (3P1/2P1)*5C5*0C0*0C0 = 3*1*1*1 = 3

Similarly,
Case-2: 401, total possible arrangements : (3P1)*5C4*1C0*1C1 = 6*5*1*1 = 30
Case-3: 302, total possible arrangements : (3P1)*5C3*2C0*2C2 = 6*10*1*1 = 60
Case-4: 311, total possible arrangements : (3P1/2P1)*5C3*2C1*1C1 = 3*10*2*1 = 60
Case-4: 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 visualize-able 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.

### GMAT/MBA Expert

GMAT Instructor
Joined
09 Apr 2015
Posted:
1461 messages
Followed by:
16 members
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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
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.

_________________
Jeffrey Miller Head of GMAT Instruction

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