Section 6-2

(24) 50 people are in a race. How many different 1st, 2nd, 3rd, 4th, and 5th place finishes can there be? (Disregard ties.)

Think of this as lining up 5 of the 50 people as 1st, 2nd, 3rd, 4th, and 5th place finishers.
There are P_{50, 5} = 50! / (50 - 5)! = 50! / 45! = (50)(49)(48)(47)(46)(45!) / 45! = 254,251,200 ways that this can happen.

(26) Nine cards are numbered 1 to 9. A hand of 3 cards is dealt. How many outcomes are there if...

(A) Order is considered. P_{9, 3} = 9! / (9-3)! = 9! / 6! = (9)(8)(7)(6!) / 6! = (9)(8)(7) = 504.

(B) Order is not considered. C_{9, 3} = 9! / (3! (9-3)!) = 9! / (3! 6!) = (9)(8)(7)(6!) / ((6)(6!)) = 84.

(32) From a 52-card deck, how many 5-card hands will have 2 clubs and 3 hearts?

Such a hand can be made with 2 operations:
O₁ : pick 2 out of 13 clubs. There are C_{13, 2} = 78 ways of doing this.
O₂ : pick 3 out of 13 hearts. There are C_{13, 3} = 286 ways of doing this.

By the multiplication principle, there are a total of (78)(286) = 22,308 such hands.

(34) Three departments have 12, 15, and 18 members, respectively. Each department selects a delegate and an alternate. In how many ways can this be done?

There are 3 operations here.
O₁ : Department #1 picks delegate and alternate; P_{12, 2} = 132 ways.
O₂ : Department #2 picks delegate and alternate; P _{15, 2} = 210 ways.
O₃ : Department #3 picks delegate and alternate; P_{18, 2} = 306 ways.

By the multiplication principle, there are a total of (132)(210)(306) = 8,482,320 ways.

(38) Five points are selected on a circle.

(A) How many chords can be drawn? C_{5, 2} = 10.

(B) How many triangles can be drawn? C_{5, 3} = 10.

(42) How many 4-person committees are possible from a group of 9 people if...

(A) There are no restrictions.
This is just C_{9, 4} = 126.

(B) Jim and Mary must be on the committee.
Two more people from the remaining 7 people must be chosen to complete the committee. This can be done in C_{7, 2} = 21 ways.

(C) Either Jim or Mary (but not both) must be on the committee.
Let A be the set of 4-member committies with Jim, but not Mary.
Let B be the set of 4-member committies with Mary, but not Jim.

We count the number of committies in A as follows. Jim is in each committee, so we need to choose 3 out of 7 people (9 minus Jim and Mary) to complete a committee. There are C_{7, 3} = 35 ways of doing this, so there are 35 committies in the set A.

For the same reason, there are 35 committees in the set B.

Now, A and B are disjoint, so the addition principle gives an answer of
n(A ∪ B) = n(A) + n(B) = 35 + 35 = 70.

Therefore, there are 70 committies that contain Jim or Mary (but not both).