A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?  Choose the correct choice from the following answers.
A) 635 B) 700 C) 1404 D) 2620 E) 3510
8 Men - 5 Women   Choose 6 ( two cases)   Case 1:2 men, 4 women   No of ways to choose 2 men out of 8 is 8C2 . This includes that one case in which the two men who refuse to serve together are included. So men can be chosen in 8C2 - 1 ways.   No. of ways to choose 4 women out of 5 is 5C4.   Number of ways of choosing 2 men and 4 women = (8C2 - 1)*(5C4) = 27*5= 
<gadget id="calculator">27*5</gadget>
<output>135</output>
135   Case 2: 3 men, 3 women   No of ways to choose 3 men out of 8 is 8C3. No of ways to choose those two men together is 6C1 (You choose them and choose one more to make 3). So men can be chosen in 8C3 - 6C1 ways.   No of ways to choose 3 women out of 5 is 5C3.   Number of ways of choosing 3 men and 3 women= (8C3 - 6C1)*(5C3) = 500   Total number of ways = 135 + 500= 
<gadget id="calculator">135 + 500</gadget>
<output>635</output>
635   635NSWER:635
<result>A</result>
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?