In how many ways 5 letters be posted in 3 post boxes,if any number of letters can be posted in all of the three post boxes??

first of all the 5 letters can be arranged in 5! ways or 120 ways

now let's place them in the boxes, assume that some

A B C
5 0 0
4 0 1
4 1 0
3 2 0
3 0 2
3 1 1
2 3 0
2 0 3
2 2 1
2 1 2
1 4 0
1 0 4
1 3 1
1 1 3
1 2 2
0 5 0
0 0 5
0 4 1
0 1 4
0 3 2
0 2 3 , I counted 21 of these, check this

number of ways = 120(21) = 2520

Answer should be 231.

Happy Sunday!

If you consider the letters first then you have
____ then ______ then _____ then _____ then _____
and each of the five letters can have three places to go
3 then 3 then 3 then 3 then 3
thus
3 x 3 x 3 x 3 x 3 = 243 possibilities
Did you have a typo when you wrote 231 at 9:04 this morning??
The correct answer is 243 : )

To solve this problem, we can use the concept of combinations with repetition.

In this scenario, we have 5 letters that can be distributed among 3 post boxes. Since each post box can receive any number of letters (including zero), we need to find the number of combinations.

To solve this, we can use the stars and bars method:

Step 1: Represent the 5 letters as stars (*****).
Step 2: Arrange two bars (|) to separate the post boxes (e.g., ||*****||).
Step 3: The number of stars to the left of the first bar represents the number of letters in the first post box, and so on.

For example, if we have ||***|*|**, it means there are 0 letters in the first post box, 3 letters in the second post box, and 2 letters in the third post box.

To find the number of combinations, we need to decide the positions of the bars among the available spaces. The number of spaces is equal to the number of letters (5 in this case) plus the number of separators (2 in this case).

We can calculate the number of combinations using the formula:

Number of combinations = (n + k - 1) choose (k - 1)

where n is the number of letters and k is the number of post boxes.

In this case, the number of combinations will be:

Number of combinations = (5 + 3 - 1) choose (3 - 1)
= 7 choose 2
= 21

Therefore, there are 21 different ways to distribute 5 letters among 3 post boxes, allowing any number of letters to be posted in each box.