A dinner has to be hosted with 5 managers.Find the number of ways in which the managers may be selected from among 9 managers,if 2 managers will not attend the meeting together.  Choose the correct choice from the following options.
A) 1.35 B) 2.None of these C) 3.12 D) 4.126 E) 5.15
9 managers, but two of them cannot attend the meeting together.   We can split it into two cases.   1. Meeting without these two managers in it. That would mean selecting 5, from the remaining 7 which is 7C5 = 21   2. Meeting with one of the two managers. select 1 manager from two, and then select 4 from the remaining 7, which is 2C1 X 7C4 = 70.   So, answer is   21 + 70= 
<gadget id="calculator">  21 + 70</gadget>
<output>91</output>
91.   Answer : 2.None of these
<result>B</result>
A dinner has to be hosted with 5 managers.Find the number of ways in which the managers may be selected from among 9 managers,if 2 managers will not attend the meeting together.