There are $4$ different letters and $3$ identical postboxes , we want to disperse them into boxes such that one of the boxes has always more than or equal to $2$ letters, the rest do not have any restriction. how many arrangements can be made ?
I found that there are $14$ ways by brutal force . However, i think that there must be combinatorial approach. To find combinatorial approach , i thought that if the boxes were different then we would have $3$ choices to select which boxes will have more than or equal to $2$ object.After that,there are $33$ ways to distribute in desired condition. As a result ,there will be $3 \times 33 =99$ choices. But there are not any relation between $14$ and $99$. Hence , i need help..
There are four types of distributions: four letters in one box; three letters in one box and one letter in a different box; two letters each in two of the boxes; two letters in one box and one letter each in each of the other boxes.
Four letters in one box: There is one way to place all four letters in one box.
Three letters in one box and the other letter in a different box: Select which three of the four letters are placed in the same box. Place those letters in a box. Place the remaining letters in one of the other boxes. There are $$\binom{4}{3} = 4$$ such distributions.
Two letters each in two of the boxes: There are three ways to match one of the other three letters with the letter whose address appears first in an alphabetical list. Place those letters in one box and place the other two letters in another box. There are $3$ such distributions.
Two letters in one box and one letter each in each of the other boxes: Select which two of the four letters will be placed in the same box. Place those letters in a box. Place one letter each in each of the other boxes. There are $\binom{4}{2} = 6$ such distributions.
Total: Since these four cases are mutually exclusive and exhaustive, there are $$1 + 4 + 3 + 6 = 14$$ possible distributions of four letters into three boxes if at least one box must contain at least two letters.
Since it is not possible to distribute four letters to three boxes without placing at least two letters in one box, our answer should equal the sum of the Stirling numbers of the second kind $S(4, 1) + S(4, 2) + S(4, 3) = 1 + 7 + 6 = 14$, which it does.