probability

A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen? 


119 

1,200 

3,240 

3,600 

14,400 

This is a combination problem, so you will be using the formula: 

nCr = (n!)/(r!*(n-r)!) 

You must break the problem up into two Cases. Find the combination of each part and then add them. 

Case 1 is selecting exactly two officers and three civilians. 

For the officers, 

5C2 = (5!)/(2!*(5-2)!) = 10 

For the civilians, 

9C3 = (9!)/(3!*(9-3)!) = 84 

So, 10*84 combinations for Case 1 = 840. 

Case 2 is selecting exactly three officers and two civilians. 

For the officers, 

5C3 = (5!)/(3!*(5-3)!) = 10 

For the civilians, 

9C2 = (9!)/(2!*(9-2)!) = 36 

So, 10*36 combinations for Case 2 = 360. 

Adding Case 1 and Case 2 = 1200. 

The answer is B.