SOLUTION: Respected Sir, my Question is: Find the number of ways in which one can post 5 letters in 7 letter boxes.I know that we can solve this by doing 7X7X7X7X7=7^5 But

Algebra ->  Permutations -> SOLUTION: Respected Sir, my Question is: Find the number of ways in which one can post 5 letters in 7 letter boxes.I know that we can solve this by doing 7X7X7X7X7=7^5 But       Log On


   

Question 734066: Respected Sir,
my Question is: Find the number of ways in which one can post 5 letters in 7 letter boxes.I know that we can solve this by doing 7X7X7X7X7=7^5
But my question is that ,why can't we do 5X5X5X5X5X5X5=5^7? Why ,this way of doing is wrong? how will you provide a valid explanation/reason for this?
Thank you in anticipation .

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The calculation done as 7X7X7X7X7=7^5 is based on the fact that for each letter you have exactly 7 choices on where to place it, so for each of the 7 choices that could be made for the first letter there are 7 choices that can be made for the second letter, and the choices multiply. There are 7X7 possibilities for where the first 2 letters will be placed, and for each of those cases, there are 7 possibilities for what to do with the third letter. For each letter there are exactly 7 choices, and the number of possible outcomes is the product of all those 7's.

Starting with the mailboxes does not lead to an answer (or at least not easily).
5X5X5X5X5X5X5=5^7 would work if there were 5 possibilities for each of the 7 mailboxes, no matter what happened with the other mailboxes, but that is not the case.
Mailbox number 1 could get no letters at all, could get one of the 5 letters or could get a set of 2 or more letters. The possibilities are many, a lot more that 5. The calculation of the number of possibilities for mailbox number 1 gets complicated, but could be done.
The real problem is that each possible outcome for mailbox number one leads to different possibilities for mailbox number 2, and the other mailboxes.
If you put no letters in mailbox number 1, there are many possibilities for the second mailbox, but if you put all 5 letters in the first mailbox, there is just one possible outcome for the other mailboxes. You would have to start making huge diagrams of all the possible outcomes, and would probably run out of paper for your calculations.