Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: counting principles [#permalink]
18 Nov 2009, 04:55

2

This post received KUDOS

Expert's post

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
_________________

Re: counting principles [#permalink]
17 Feb 2010, 02:47

Ravshonbek wrote:

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?

warm up

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

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: counting principles [#permalink]
10 Feb 2012, 03:03

Bunuel wrote:

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

Hi Bunuel,

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

******************** Push+1 kudos button please, if you like my post.

Re: In how many different ways can trhee letters be posted from [#permalink]
10 Feb 2012, 04:05

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

Re: counting principles [#permalink]
10 Feb 2012, 08:13

1

This post received KUDOS

Expert's post

mohankumarbd wrote:

Bunuel wrote:

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

Hi Bunuel,

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

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.

Re: counting principles [#permalink]
25 May 2013, 07:22

Bunuel wrote:

mohankumarbd wrote:

Bunuel wrote:

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

Hi Bunuel,

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

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.

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

Re: counting principles [#permalink]
25 May 2013, 12:06

Quote:

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

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
_________________