The students S_{1}, S_{2}, ldots, S_{10} are | vedclass

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(D) Let $n(C)$ be the number of students in group $C$. Since each group must have at least one student,the remaining $10 - n(C)$ students must be dist

Vedclass · Accepted Answer

$31650$

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(D) Let $n(C)$ be the number of students in group $C$. Since each group must have at least one student,the remaining $10 - n(C)$ students must be distributed into groups $A$ and $B$ such that each of $A$ and $B$ has at least one student.For a fixed set of $n(C)$ students in group $C$,the number of ways to distribute the remaining $10 - n(C)$ students into groups $A$ and $B$ such that each is non-empty is $2^{10-n(C)} - 2$.Case $1$: $n(C) = 1$.Number of ways $= {}^{10}C_{1} 	imes (2^{9} - 2) = 10 	imes (512 - 2) = 10 	imes 510 = 5100$.Case $2$: $n(C) = 2$.Number of ways $= {}^{10}C_{2} 	imes (2^{8} - 2) = 45 	imes (256 - 2) = 45 	imes 254 = 11430$.Case $3$: $n(C) = 3$.Number of ways $= {}^{10}C_{3} 	imes (2^{7} - 2) = 120 	imes (128 - 2) = 120 	imes 126 = 15120$.Total number of possibilities $= 5100 + 11430 + 15120 = 31650$.

The students $S_{1}, S_{2}, \ldots, S_{10}$ 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 ........ .

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