Question

The students $S_1,S_2,..........S_{10}$ are to be divided into $3$ groups $A,B$ and $C$ such that each group has at least one student and the group $C$ has at most $3$ students. Then the total number of possibilities of forming such groups is

Answer 1

Step 1: Determine the different cases

Given, number of students $=10$.

There are $3$ cases satisfying the given conditions.

Case 1: Group $C$ has $1$ student and other are filled in $A,B$

Case 2: Group $C$ has $2$ student and other are filled in $A,B$

Case 3: Group $C$ has $3$ student and other are filled in $A,B$

Step 2: Find the number of possibilities in all the three cases

Case 1: Number of possibilities$={}^{10}{C}_1\left(2^9-2\right)$ , where $2^9$ is the possibility of filling in $A,B$ and $-2$ is the possibility of $A$ and $B$ having no student.

Case 2: Number of possibilities$={}^{10}{C}_2\left(2^8-2\right)$ , where $2^8$ is the possibility of filling in $A,B$ and $-2$ is the possibility of $A$ and $B$ having no student.

Case 3: Number of possibilities$={}^{10}{C}_3\left(2^7-2\right)$ , where $2^7$ is the possibility of filling in $A,B$ and $-2$ is the possibility of $A$ and $B$ having no student.

Step 3: Find the total number of possibilities

Total number of possibilities$={}^{10}{C}_1(2^9-2)+{}^{10}{C}_2(2^8-2)+{}^{10}{C}_3(2^7-2)$

$$\begin{gathered} =\frac{10!}{9!\times 1!}(2^9-2)+\frac{10!}{8!\times 2!}(2^8-2)+\frac{10!}{7!\times 3!}(2^7-2) \\ =10(2^9-2)+\frac{10\times 9}{2}(2^8-2)+\frac{10\times 9\times 8}{3\times 2}(2^7-2) \\ =10\times 2^9-20+2^8\times 45-90+120\times 2^7-240 \\ =2^7\left(40+90+120\right)-20-90-240 \\ =128\left(250\right)-350 \\ =31650 \end{gathered}$$

Hence, the total number of possibilities $=31650$.