Page 178 - DMTH202_BASIC_MATHEMATICS_II

Unit 13: Combinations

Thus we have 8×330 = 2640 ways to select a 5-member committee with exactly 4 female members Notes
from this club.
At least 4 females?

Solution: We found that there are 2640 ways to select exactly 4 females for our 5-member
committee. If we need at least 4 females, we could have 4 females or 5 females, couldn’t we?
How many ways can we select 5 females from the 11 females in the club, given that order doesn’t
matter?
C(11, 5) = 11!/[(11 – 5)! ×5!] = 11!/(6! ×5!) = (11×10×9×8×7)/ (5×4×3×2×1) = 462 ways.

So to have at least 4 females, we could have C(11, 4) ×C(8, 1) or C(11, 5). Since there are 330 ways
to achieve the first and 462 ways to achieve the second, we have
330 + 462 = 792 ways to select a 5-member committee with at least 4 female members from the club.

No more than 2 males?
Solution: “No more than 2” males means we could have 1 male or 2 males on our 5-member
committee. Since we have the condition of the 5-member committee, we have the following
possibilities: (1) 1 male and 4 females, or (2) 2 males and 3 females. Since order doesn’t matter,
we have C(8, 1)C(11, 4)  C(8, 2)C(11, 3) ways to select the committee. Applying the combinations
formulas, we have {8!/[(8 – 1)! 1!]}  {11!/[(11 – 4)! 4!]} + {8!/[(8 – 2)! 2!]} × {11!/[(11 – 3)! ×3!]} =
(8)(330) + (28)(165) = 2540 + 4620 = 7260 ways to select a 5-member committee with no more than
2 male members from the club.

Example: Suppose we have the same club with 8 males and 11 females, but this time we
want to know how many committees could be chosen with at least 1 but no more than
3 members, regardless of gender. Therefore, committees must have 1, 2, or 3 members. What
are the possible committee memberships?

Solution: The possibilities are 1M or 1F or 2M or 2F or 1M & 1F or 1M & 2F or 2M & 1F or 3M or
3F. Does order matter? No, it does not. So we use our combinations formulas again for each
possible committee make-up and add them to get the solution to our problem.
C(8, 1) + C(11, 1) + C(8, 2) + C(11, 2) + C(8, 1)C(11, 1) + C(8, 1) C(11, 2) + C(8, 2) C(11, 1) + C(8, 3)
+ C(11, 3).

Example: In a discussion of 9 schools, how many intraconference football games are
played throughout the period if the teams all play each other exactly once?
When the teams plays with each other, order is not an issue, we are counting match ups. For each
game there is a collection of two teams playing. So we can utilize combinations to assist us out here.
Observe that if we were putting these teams in any sort of order, then we would require using
permutations to explain the problem.
But here, order is not an issue, so we will use combinations.
First we are required to find n and r :

If n is the number of teams we have to select from, what do you think n here?
If you said n = 9 you are correct!!! There are 9 teams in the conference.
If r is the number of teams we are using at a time, what do you think r is?

If you said r = 2, pat yourself on the back!! 2 teams play per game.

LOVELY PROFESSIONAL UNIVERSITY 173