How many ways are there to distribute $15$ different letters into $5$ post boxes if:
a-) Post boxes are different and each will have $3$ letters.
b-) Post boxes are identical and each will have $3$ letters.
c-) Post boxes are identical and they will have $3,3,2,2,5$ letters.
d-) Post boxes are identical and they will have $3,1,4,2,5$ letters.
I want you to check my thinking way. Is it true or not ?
My solution :
a-) It is easy we can find it by saying choose $3$ elements for the first box , choose $3$ elements for the second box so on. Then , the answer is $C(15,3) \times C(12,3) \times C(9,3) \times C(6,3) \times C(3,3) $.
b-) Lets show the cluster of letters which have the same number of letter with red balls. Then we have $5$ identical red balls representing cluster of $3$ letters. To solve question easily , i assume that post boxes are different. Then , we can disperse $5$ identical objects to $5$ different post boxes with each boxes have one red ball by $1$ way.
So , we can write $1 \times C(15,3) \times C(12,3) \times C(9,3) \times C(6,3) \times C(3,3) $. However , the post boxes were identical , so i should multiply it with $1/5!$
Answer : $\frac{1}{5!} \times 1 \times C(15,3) \times C(12,3) \times C(9,3) \times C(6,3) \times C(3,3)$
c-)Lets firstly assume that the post boxes are different .Moreover , show the cluster of $3$ letters with red balls and the cluster of $2$ letters with blue balls and the cluster of $5$ letters with green balls.
Now ,we have $2$ red balls , $2$ blue balls , $1$ green balls to distribute $5$ different boxes. We can do it $\frac {5!}{2! \times 2! \times 1!}$ different ways.
Then , $\frac {5!}{2! \times 2! \times 1!} \times C(15,3) \times C(12,3) \times C(9,2) \times C(7,2) \times C(5,5)$
However , in original , the boxes were identical but we assumed them as different to solve easily. Then , we should multiply with $\frac{1}{5!}$ to make them identitical $\color{blue}{again}$.
So, the real answer is $\frac{1}{5!} \times \frac {5!}{2! \times 2! \times 1!} \times C(15,3) \times C(12,3) \times C(9,2) \times C(7,2) \times C(5,5)$
d-)Lets firstly assume that the post boxes are different. Moreover , because of all clusters have different number of letters , lets show them $5$ different colors.
Then , we can disperse $5$ different objects to $5$ different boxes by $5!$ ways. Hence ,we can write $5! \times C(15,3) \times C(12,1) \times C(11,4) \times C(7,2) \times C(5,5)$.
However , the post boxes were identical ,so i should multiply with $\frac{1}{5!}$ to make them identitical $\color{blue}{again}$.
Then , the answer is $C(15,3) \times C(12,1) \times C(11,4) \times C(7,2) \times C(5,5)$.
Is my thinking $\color{red}{WAY}$ correct ? Can i convert diffrent object into identical objects like i did in my solutions. Thank you..