In how many different ways can trhee letters be posted from

Question

1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?

2. what if there is no restriction, that is, if two or more letters can be posted from the same box?

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Bunuel · Accepted Answer

1. In how many different ways can trhee letters be posted from seven different postboxes assuming no two letters can be posted from the same postbox?

First letter could be sent from ANY of the seven postboxes - 7 (7 options);
Second letter could be sent from the SIX postboxes left - 6 (6 options);
Third letter could be sent from the FIVE postboxes left - 5 (5 options);

Total # of ways =7*6*5=210

2. what if there is no restriction, that is, if two or more letters can be posted from the same box?

In this problem we don't have restriction, thus ANY letter could be sent from ANY postboxes =7*7*7=7^3=343

jeeteshsingh · Answer

1. 7 x 6 x 5 = 210
2. 7 x 7 x 7 = 343

GetThisDone · Answer

Hi Bunuel,

Could you please elaborate on the second question. Couldn't figure out why.

MSoS · Answer

1. 7 (no restriction) * 6 (can not be the same as the first one) * 5 (can not be the same as the first and second one) = 210
2. 7 (no restriction) * 7 (no restriction) * 7 (no restriction) = 343

Bunuel · Answer

Welcome to GMAT Club. Please find below answer to your question:

"Two or more letters can be posted from the same box" means that all 3 letters can be posted from the same postbox (so we don't have the restriction we had for the first question).

Now, since there are 7 postboxes then each of these 3 letters has 7 options to be posted from, total # of ways is 7*7*7=7^3.

Hope it's clear.

cumulonimbus · Answer

Hi Bunnel,

I tried to do the second question via combinatorics, but i am not able to figure it out, please check the below method and guide where i went wrong

= all three in one box +2 in one box and the last one in a different box + all three in different boxes
= 3c3*7+3c2*7c1*6c5+3c1*7c3
= 7+ 3*7*6+3*7*6*5
= 7 + 126 + 270 = wrong

BangOn · Answer

Think it in this way,
First letter can go to any 7 post offices
Same case with second and same case with the third letter as well so 7*7*7

icetray · Answer

Could someone please tell me why part 1 cannot be answered using "7C3" = 35 ways?

Many thanks.

EMPOWERgmatRichC · Answer

Hi icetray,

Since the question is worded in a "vague" way, you bring up an interesting interpretation of it. Thankfully, questions on the Official GMAT are worded to remove ambiguity and "bias" on the part of the reader, so you won't have to worry about that on Test Day. This prompt reads as if it were created by the original poster, so it's not clear what he/she was "intending" the question to mean.

As it is, your interpretation of the prompt makes a lot of sense - there does not seem to be any reason why we should emphasize the "order" of the letters (there's no reference to "first letter", "second letter", "third letter" and no reference to "arrangements"). Using post-boxes ABC would be same as BCA, CBA, etc., so if we interpret the prompt as a "combinations" question, then 7c3 = 35 would be correct.

GMAT assassins aren't born, they're made,
Rich

cptholt · Answer

3. What if we assume no three letters can be posted from same postbox?

bv8562 · Answer

Hi Banuel,

I tried to do it this way:

(7C1 x 3C1) x (6C1 x 2C1) x (5C1 x 1C1) = 1260

Could you please tell me what am I doing wrong here?

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In how many different ways can trhee letters be posted from

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