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

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mensanumber: Senior | Next Rank: 100 Posts; Posts: 31; Joined: Wed Oct 23, 2013 6:43 amFollowed by: 1 member; Thanked: 2 times; GMAT Score:750

by mensanumber » Wed Mar 07, 2018 8:18 am

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

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Source: Beat The GMAT — Problem Solving | Wed Mar 07, 2018 8:18 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Wed Mar 07, 2018 8:26 am

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 (= 3â�µ ways)

Answer: D
--------------------------

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mensanumber: Senior | Next Rank: 100 Posts; Posts: 31; Joined: Wed Oct 23, 2013 6:43 amFollowed by: 1 member; Thanked: 2 times; GMAT Score:750

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

by mensanumber » Wed Mar 07, 2018 8:28 am

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

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Wed Mar 07, 2018 9:07 am

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

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regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Wed Mar 07, 2018 9:08 am

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

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mensanumber: Senior | Next Rank: 100 Posts; Posts: 31; Joined: Wed Oct 23, 2013 6:43 amFollowed by: 1 member; Thanked: 2 times; GMAT Score:750

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

by mensanumber » Wed Mar 07, 2018 9:58 am

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,
Thanks for your reply.

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

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mensanumber: Senior | Next Rank: 100 Posts; Posts: 31; Joined: Wed Oct 23, 2013 6:43 amFollowed by: 1 member; Thanked: 2 times; GMAT Score:750

by mensanumber » Wed Mar 07, 2018 10:15 am

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

Thanks for your reply Regor.

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

Like this post

regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Wed Mar 07, 2018 11:34 am

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
Thanks for your reply Regor.

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.

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 amFollowed by: 22 members; Location: New York, NY; Thanked: 39 times

by Jeff@TargetTestPrep » Mon Mar 12, 2018 10:31 am

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.

Answer: D

Jeffrey Miller
Head of GMAT Instruction
[email protected]

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

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rohit56: Senior | Next Rank: 100 Posts; Posts: 35; Joined: Sat Dec 15, 2018 12:06 am

by rohit56 » Mon Dec 17, 2018 10:07 pm

Each letter has 3 options.
So, total no of ways = 3 x 3 x 3 x 3 x 3 = 3^5

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