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

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RBBmba@2014 GMAT Destroyer! Default Avatar
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In how many ways can a person post 5 letters in 4 letter box Post Mon Mar 09, 2015 10:44 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    Post 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 (= 4⁵ ways)

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

    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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    Cheers,
    Brent

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    madhusudhan237 Just gettin' started! Default Avatar
    Joined
    30 Oct 2015
    Posted:
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    Post 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]

    Post 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

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    emilytay23 Just gettin' started!
    Joined
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    Posted:
    2 messages
    Post 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. Sad

    Please help!

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