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the 5 letter boxes

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by sanju09 » Fri Mar 23, 2012 3:04 am
In how many ways can 8 letters be posted into 5 letter boxes if any number of the 8 letters can be posted in any of the 5 letter boxes?
(A) 8! / (3! 5!)
(B) 8! /3!
(C) 5^6
(D) 8^5
(E) 5^8




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by Brent@GMATPrepNow » Fri Mar 23, 2012 6:03 am
sanju09 wrote:In how many ways can 8 letters be posted into 5 letter boxes if any number of the 8 letters can be posted in any of the 5 letter boxes?
(A) 8! / (3! 5!)
(B) 8! /3!
(C) 5^6
(D) 8^5
(E) 5^8
Please note: I am assuming that each letter is unique and each letter box is unique.

We can take the task of posting the 8 letters and break it into stages.
Stage 1: Determine which box the first letter goes into.
Stage 2: Determine which box the 2nd letter goes into.
Stage 3: Determine which box the 3rd letter goes into.
. . . etc.

Stage 1: There are 5 boxes to choose from, so we can accomplish stage 1 in 5 ways.
Stage 2: There are 5 boxes to choose from, so we can accomplish stage 2 in 5 ways.
Stage 3: There are 5 boxes to choose from, so we can accomplish stage 3 in 5 ways.
. . . etc.

So, the total number of ways to accomplish all 8 stages (and place all 8 letters) = (5)(5)(5)(5)(5)(5)(5)(5)
= (5)^8 = E

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by icemanKK » Fri Mar 23, 2012 6:10 pm
Hi Brent,

I have come across logic that you posted in a lot of places but cant seem to figure it out properly.

I shall provide an example below. Can you please tell me if this scenario and it solution is comparable to the above problem.

Lets us say we have 3 letters and 2 places to put them in (similar to the problem above)

The number of ways the first place can be filled = 3
The number of ways the second place can be filled = 2 (problem is without replacement as in the case of the letters)

Total No. of ways = 6 but generally I find that these have solutions 3^2.

I cant seem to differentiate between the 2 types of problems (the one posted above and mine).
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by sanju09 » Wed Mar 28, 2012 11:30 pm
icemanKK wrote:Hi Brent,

I have come across logic that you posted in a lot of places but cant seem to figure it out properly.

I shall provide an example below. Can you please tell me if this scenario and it solution is comparable to the above problem.

Lets us say we have 3 letters and 2 places to put them in (similar to the problem above)

The number of ways the first place can be filled = 3
The number of ways the second place can be filled = 2 (problem is without replacement as in the case of the letters)

Total No. of ways = 6 but generally I find that these have solutions 3^2.

I cant seem to differentiate between the 2 types of problems (the one posted above and mine).
Your example is perhaps not the same as the one posted here. Although the wordings in your example are not revealing, but I can still assume that you mean only 1 letter can be posted in one letter box. In that case, your take is correct. But, in the same situation, had it been mentioned that any number of the 3 letters can be posted in any of the two boxes, the case would have been same as this thread and the answer would have been 3^2.
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by vikram4689 » Mon Apr 09, 2012 10:28 pm
Hi ICEMAN,

As Sanjay mentioned, In your case only letter is allowed in 1 box hence. Total no of permuations =6
ab
ba
ac
ca
cb
bc

However, if any number of letters were allowed (as in original problem)
Total number of cases will be = 2*2*2 = 2^3 NOT 3^2 as Each letter has 2 options for letter box.

Check out this helpful link https://www.mathsisfun.com/combinatorics ... tions.html
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by GMATGuruNY » Tue Apr 10, 2012 4:16 am
icemanKK wrote:Hi Brent,

I have come across logic that you posted in a lot of places but cant seem to figure it out properly.

I shall provide an example below. Can you please tell me if this scenario and it solution is comparable to the above problem.

Lets us say we have 3 letters and 2 places to put them in (similar to the problem above)

The number of ways the first place can be filled = 3
The number of ways the second place can be filled = 2 (problem is without replacement as in the case of the letters)

Total No. of ways = 6 but generally I find that these have solutions 3^2.

I cant seem to differentiate between the 2 types of problems (the one posted above and mine).
How many ways can 3 letters be placed into 2 boxes, if it is possible for all 3 letters to be placed into the same box?

EACH LETTER must choose a box because each letter must be placed inside one of the boxes.
But it is NOT necessary that EACH BOX choose a letter, since it's OK for a box to be empty.
Thus, we count the number of options from the viewpoint of EACH LETTER.
The number of options for the first letter = 2. (Either of the 2 boxes.)
The number of options for the second letter = 2. (Either of the 2 boxes.)
The number of options for the third letter = 2. (Either of the 2 boxes.)
To combine these options, we multiply:
2*2*2 = 2³ = 8.
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