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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.If the team size is $r$,then $r \ge 3$ and $(10 - r) \ge 3$.From $(10 - r) \ge 3$,we get $r \le 7$.Thus,the possible values for the team size $r$ are $3, 4, 5, 6, 7$.The number of ways to form the team is the sum of combinations for these sizes:Total ways $= {^{10}C_3} + {^{10}C_4} + {^{10}C_5} + {^{10}C_6} + {^{10}C_7}$.We know that the sum of all binomial coefficients is $\sum_{k=0}^{10} {^{10}C_k} = 2^{10} = 1024$.We can write: $\sum_{k=3}^{7} {^{10}C_k} = 2^{10} - ({^{10}C_0} + {^{10}C_1} + {^{10}C_2} + {^{10}C_8} + {^{10}C_9} + {^{10}C_{10}})$.Since ${^{10}C_0} = {^{10}C_{10}} = 1$,${^{10}C_1} = {^{10}C_9} = 10$,and ${^{10}C_2} = {^{10}C_8} = 45$,Total ways $= 1024 - 2 	imes (1 + 10 + 45) = 1024 - 2 	imes (56) = 1024 - 112 = 912$.

$A$ 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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