Solve the following permutation question.

Question

Suppose that a department contains 8 men and 16 women. How many ways are there to form a committee with 6 members if it must have strictly more women than men?

Answer 1

Thinking about this particular problem, it is easier to consider "fixed" chair members and then "flexible" members.  Lets start with the basic idea of MALE to FEMALE possibilities:

 

• 4 women, 2 men [case A]
• 5 women, 1 man [case B]
• 6 women, 0 men [case C]

Notice that 3 women would not be possible since there would then be 3 men and the problem requirement says that there must be STRICTLY more women then men.

 

Since the overall committee can be formed as "CASE A  or   CASE B" that meas the total possible ways  [case A] could be created  should be added to the total possible ways [case B] can be created.

 

In each case, the women can be placed on the committee and then the men, resulting in a combination of women and a combination of men (since the order members are picked in doesn't really matter).  This means that in EACH case, the total number of ways to select the women AND THEN men should be multiplied to find the total possible outcome for the given case

• (4 women, 2 men)=  16C4 * 8C2 [16 women choose 4  AND 8 men choose 2]
• (5 women, 1 man)=  16C5 * 8C1 [16 women choose 5  AND 8 men choose 1]
• (6 women, 0 men)=  16C6 * 8C0 [16 women choose 6  AND 8 men choose 0]**

** Notice that in the last case it could have also been just 16C6 since 8C0 will be equivalent to 1.

 

 

Taking  all of the total outcomes together would result in

 

16C4 * 8C2 + 16C5 * 8C1 + 16C6 * 8C0 = Total Possible Committees

 

Happy counting!