In how many ways can a person post 5 letters in 4 letter box

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RBBmba@2014: Legendary Member; Posts: 944; Joined: Wed May 30, 2012 8:21 amFollowed by: 5 members; Thanked: 8 times

by RBBmba@2014 » Mon Mar 09, 2015 10:44 am

In how many ways can a person post 5 letters in 4 letter boxes ?

(A) 120
(B) 600
(C) 2400
(D) 4^5
(E) 5^5

OA: D

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Source: Beat The GMAT — Problem Solving | Mon Mar 09, 2015 10:44 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 » Mon Mar 09, 2015 10:50 am

RBBmba@2014 wrote:In how many ways can a person post 5 letters in 4 letter boxes ?

(A) 120
(B) 600
(C) 2400
(D) 4^5
(E) 5^5

OA: D

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 4 available boxes, so we can complete stage 1 in 4 ways

Stage 2: Select a box for the 2nd letter to go into.
There are 4 available boxes, so we can complete stage 2 in 4 ways

Stage 3: Select a box for the 3rd letter to go into.
There are 4 available boxes, so we can complete stage 3 in 4 ways

Stage 4: Select a box for the 4th letter to go into.
There are 4 available boxes, so we can complete stage 4 in 4 ways

Stage 5: Select a box for the 5th letter to go into.
There are 4 available boxes, so we can complete stage 5 in 4 ways

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus distribute all 5 letters) in (4)(4)(4)(4)(4) ways ([spoiler]= 4â�µ ways[/spoiler])

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

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madhusudhan237: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Fri Oct 30, 2015 12:35 pm

by madhusudhan237 » Wed Dec 23, 2015 5:10 am

HI Brent

In your post, you have considered like: for each letter to get posted, there are 4 letter boxes available.

But, why not in such a way that - I have 4 letter boxes available & 5 letters to post?

In that case,

---------- ---------- ----------- -----------
Letter box-1 Letter box-2 Letter box-3 Letter box-4

So, to post 5 letters

Case1: Letter box-1 has 5 ways (all 5 letters available)
Case2: Letter box-2 has 4 ways (since, only 4 letters are available, one already posted in letter box-1)

Similarly,
Case3, Case4 will have 3, 2 ways

Total (by FCP rule) = 5 x 4 x 3 x 2 = 120 ways

I know, I am wrong! But, I am trying to get into the grove of right thinking

Thanks for your clarification!

Madhu

[/list]

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by Brent@GMATPrepNow » Thu Dec 24, 2015 6:55 am

madhusudhan237 wrote:HI Brent

In your post, you have considered like: for each letter to get posted, there are 4 letter boxes available.

But, why not in such a way that - I have 4 letter boxes available & 5 letters to post?

In that case,

---------- ---------- ----------- -----------
Letter box-1 Letter box-2 Letter box-3 Letter box-4

So, to post 5 letters

Case1: Letter box-1 has 5 ways (all 5 letters available)
Case2: Letter box-2 has 4 ways (since, only 4 letters are available, one already posted in letter box-1)

Similarly,
Case3, Case4 will have 3, 2 ways

Total (by FCP rule) = 5 x 4 x 3 x 2 = 120 ways

I know, I am wrong! But, I am trying to get into the grove of right thinking

Thanks for your clarification!

Madhu

[/list]

Your solution assumes that each letter box must one letter.
To avoid this problem, try to be very clear what is happening with each stage.
In your solution, you write: Case1: Letter box-1 has 5 ways
What does this mean?

If you had written, "stage 1: select a letter to go into mailbox 1," you might have seen the problem with this approach, because there's nothing that says mailbox 1 must contain a letter.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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emilytay23: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Thu Aug 25, 2016 7:48 pm

by emilytay23 » Thu Aug 25, 2016 7:55 pm

To anyone who could help clarify something for me:

I approached this question in the following way:

Assuming that each letter is represent by L => 5 letters = L L L L L
and
Assuming each letter box is separated by *

One way of posting 5 letters into 4 letter boxes could be represented by LL*L*L*L (2,1,1,1) or LLLL* * *L (4,0,0,1) or L*LL*LL* (1,2,2,0)
Since there are 5 identical L's and 3 identical *'s,
there are 8!/(5!3!) ways = 56 ways to post 5 letters into 4 letter boxes.

Why is this calculation different from the OA of 4^5?
Is it because in Brent's answer, it is assumed that each letter is NOT identical and each letter box is also NOT identical?

If so, what is a general rule to calculate how many ways to distribute N objects into X bins for when:
i. N objects are distinct and X bins are distinct
ii. N objects are identical and X bins are identical
iii. N objects are distinct and X bins are identical
iv. N objects are identical and X bins are distinct

?

Sorry for the long question but I have been trying to wrap my brain around this but to no avail.

Please help!

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