Q
In how many ways can 5 letters be posted in 4 mail boxes , any letter can be posted in any mail box ??
How i tried to solve
5 letters each letter has 4 choices so 4*4*4*4*4 = 4^5 ways
but I get confused, when I tried to solve the same in this way
4 mail boxes each can accept 5 letters so 5*5*5*5 = 5^4 ways.
please help me out with the basic difference between the two approaches and secondly how to solve the same ??
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a basic combination -- help needed
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GMATGuruNY
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There are fewer mailboxes than letters.smanstar wrote:Q
In how many ways can 5 letters be posted in 4 mail boxes , any letter can be posted in any mail box ??
How i tried to solve
5 letters each letter has 4 choices so 4*4*4*4*4 = 4^5 ways
but I get confused, when I tried to solve the same in this way
4 mail boxes each can accept 5 letters so 5*5*5*5 = 5^4 ways.
please help me out with the basic difference between the two approaches and secondly how to solve the same ??
The result: while not every mailbox must contain a letter, EVERY LETTER MUST CHOOSE A MAILBOX.
Thus, we count from the perspective of the LETTERS.
For each of the five letters, there are 4 choices of mailbox:
4*4*4*4*4 = 4�.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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GMATGuruNY wrote:There are fewer mailboxes than letters.smanstar wrote:Q
In how many ways can 5 letters be posted in 4 mail boxes , any letter can be posted in any mail box ??
How i tried to solve
5 letters each letter has 4 choices so 4*4*4*4*4 = 4^5 ways
but I get confused, when I tried to solve the same in this way
4 mail boxes each can accept 5 letters so 5*5*5*5 = 5^4 ways.
please help me out with the basic difference between the two approaches and secondly how to solve the same ??
The result: while not every mailbox must contain a letter, EVERY LETTER MUST CHOOSE A MAILBOX.
Thus, we count from the perspective of the LETTERS.
For each of the five letters, there are 4 choices of mailbox:
4*4*4*4*4 = 4�.
thanks for the reply but I have a doubt again
If in a problem we have to distribute 10 coins to 3 children then it should be 3*3*3...*3 = 3^10 but I am not getting the answer .
Can you please also explain whats the difference between the two problems
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GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
How many ways can 10 different coins be distributed among 3 children?smanstar wrote:GMATGuruNY wrote:There are fewer mailboxes than letters.smanstar wrote:Q
In how many ways can 5 letters be posted in 4 mail boxes , any letter can be posted in any mail box ??
How i tried to solve
5 letters each letter has 4 choices so 4*4*4*4*4 = 4^5 ways
but I get confused, when I tried to solve the same in this way
4 mail boxes each can accept 5 letters so 5*5*5*5 = 5^4 ways.
please help me out with the basic difference between the two approaches and secondly how to solve the same ??
The result: while not every mailbox must contain a letter, EVERY LETTER MUST CHOOSE A MAILBOX.
Thus, we count from the perspective of the LETTERS.
For each of the five letters, there are 4 choices of mailbox:
4*4*4*4*4 = 4�.
thanks for the reply but I have a doubt again
If in a problem we have to distribute 10 coins to 3 children then it should be 3*3*3...*3 = 3^10 but I am not getting the answer .
Can you please also explain whats the difference between the two problems
Here, the coins are all DIFFERENT.
Thus, it matters whether Coin A goes to child 1, child 2, or child 3.
In this case, the number of options for each coin = 3. (Any of the 3 children.)
To combine the number of options for each coin, we multiply:
3*3*3*3*3*3*3*3*3*3 = 3¹�.
How many ways can 10 identical coins be distributed among 3 children?
Here, the coins are all IDENTICAL.
Thus, it DOESN'T matter whether Coin A goes to child 1, child 2, or child 3.
All that matters is HOW MANY coins each child receives.
For this sort of problem, I recommend the SEPARATOR method.
Check here:
https://www.beatthegmat.com/combinations-t120668.html
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
