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(B) Let $n = 10$ be the total number of employees.We need to form a team of size $r$ such that the team includes at least $3$ employees and excludes a

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$912$

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(B) Let $n = 10$ be the total number of employees.We need to form a team of size $r$ such that the team includes at least $3$ employees and excludes at least $3$ employees.This means the number of employees in the team $r$ must satisfy $3 \le r \le 10 - 3$,which simplifies to $3 \le r \le 7$.The number of ways to form such a team is the sum of combinations $\sum_{r=3}^{7} {^{10}C_r}$.We know that $\sum_{r=0}^{10} {^{10}C_r} = 2^{10} = 1024$.The required sum is $2^{10} - ({^{10}C_0} + {^{10}C_1} + {^{10}C_2} + {^{10}C_8} + {^{10}C_9} + {^{10}C_{10}})$.Calculating the values: ${^{10}C_0} = 1$,${^{10}C_1} = 10$,${^{10}C_2} = \frac{10 	imes 9}{2} = 45$.Since ${^{10}C_8} = {^{10}C_2} = 45$,${^{10}C_9} = {^{10}C_1} = 10$,and ${^{10}C_{10}} = {^{10}C_0} = 1$.The sum of excluded terms is $(1 + 10 + 45 + 45 + 10 + 1) = 112$.Therefore,the number of ways is $1024 - 112 = 912$.

$A$ company has $10$ employees. The company has decided to form a team including at least three employees and also excluding at least three employees. Then the number of ways of forming the team is

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