IMPORTANT: The question doesn't specify whether the letters and the letter boxes are UNIQUE, but I'm going to assume that they are unique.mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
Apparently, the solution is quite simple here:mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
If that's your final solution, then you appear to be treating the 5 letters as 5 IDENTICAL letters, and you are treating the 3 mailboxes as IDENTICAL mailboxesmensanumber wrote:Apparently, the solution is quite simple here:mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
each letter will have 3 boxes to choose from and so answer 3^5.
But I am having a tough time visualizing this solution. The way I am looking at is:
1box 2box 3box
5 0 0
4 0 1
3 0 2
3 1 1
2 2 1
Now these can be arranged internally for example - 500 could be 005 etc.
What mistake am I making? Thanks
Not making any mistakes so far but haven't completed the thought.mensanumber wrote:Apparently, the solution is quite simple here:mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
each letter will have 3 boxes to choose from and so answer 3^5.
But I am having a tough time visualizing this solution. The way I am looking at is:
1box 2box 3box
5 0 0
4 0 1
3 0 2
3 1 1
2 2 1
Now these can be arranged internally for example - 500 could be 005 etc.
What mistake am I making? Thanks
Hi Brent,Brent@GMATPrepNow wrote:If that's your final solution, then you appear to be treating the 5 letters as 5 IDENTICAL letters, and you are treating the 3 mailboxes as IDENTICAL mailboxesmensanumber wrote:Apparently, the solution is quite simple here:mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
each letter will have 3 boxes to choose from and so answer 3^5.
But I am having a tough time visualizing this solution. The way I am looking at is:
1box 2box 3box
5 0 0
4 0 1
3 0 2
3 1 1
2 2 1
Now these can be arranged internally for example - 500 could be 005 etc.
What mistake am I making? Thanks
Cheers,
Brent
Thanks for your reply Regor.regor60 wrote:Not making any mistakes so far but haven't completed the thought.mensanumber wrote:Apparently, the solution is quite simple here:mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
each letter will have 3 boxes to choose from and so answer 3^5.
But I am having a tough time visualizing this solution. The way I am looking at is:
1box 2box 3box
5 0 0
4 0 1
3 0 2
3 1 1
2 2 1
Now these can be arranged internally for example - 500 could be 005 etc.
What mistake am I making? Thanks
As you suggest, 500 can be arranged 3 ways.
Likewise, 401 can be arranged 3 ways.
And so on. All together there are then 3x3x3x3x3 = 3^5 ways
I read your response too quickly so ignore what I wrote, it's not correct. The way you've done it is correct.mensanumber wrote:Thanks for your reply Regor.regor60 wrote:Not making any mistakes so far but haven't completed the thought.mensanumber wrote:Apparently, the solution is quite simple here:mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
each letter will have 3 boxes to choose from and so answer 3^5.
But I am having a tough time visualizing this solution. The way I am looking at is:
1box 2box 3box
5 0 0
4 0 1
3 0 2
3 1 1
2 2 1
Now these can be arranged internally for example - 500 could be 005 etc.
What mistake am I making? Thanks
As you suggest, 500 can be arranged 3 ways.
Likewise, 401 can be arranged 3 ways.
And so on. All together there are then 3x3x3x3x3 = 3^5 ways
Here is my full solution:
1box 2box 3box
5 0 0......................................Case-1
4 0 1......................................Case-2
3 0 2......................................Case-3
3 1 1......................................Case-4
2 2 1......................................Case-5
Let's consider each case individually,
Case-1
500 i.e. 5 letters in box1, 0 letters in box-2, 0 letters box-3. Now, there will be 3 such scenarios, 3P1/2P1 = 3. (by MISSISSIPPI rule dividing by 2P1 for two identical zeros) These 3 scenarios are 500, 050, 005
Since letters distinct, for each of these scenarios: 5C5*0C0*0C0
So, total possible arrangements for case-1 : (3P1/2P1)*5C5*0C0*0C0 = 3*1*1*1 = 3
Similarly,
Case-2: 401, total possible arrangements : (3P1)*5C4*1C0*1C1 = 6*5*1*1 = 30
Case-3: 302, total possible arrangements : (3P1)*5C3*2C0*2C2 = 6*10*1*1 = 60
Case-4: 311, total possible arrangements : (3P1/2P1)*5C3*2C1*1C1 = 3*10*2*1 = 60
Case-4: 221, total possible arrangements : (3P1/2P1)*5C2*3C2*1C1 = 3*10*3*1 = 90
Total of all cases = 3+30+60+60+90 = 243 which is indeed 3^5
But what irritates me is my long winded solution is visualize-able to me and not your more elegant solution.
Can you help me see it from your point of view?
Thanks again
Since each letter can be put into any of the 3 boxes, then each letter has 3 choices. Thus, the number of ways a person can put 5 letters in 3 boxes is 3 x 3 x 3 x 3 x 3 = 3^5.mensanumber wrote:In how many ways can a person post 5 letters in 3 letter boxes?
(A) 480
(B) 1024
(C) 54
(D) 3^5
(E) 5^3
