**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:

O_{1} : pick 2 out of 13 clubs. There are C_{13, 2 }=
78 ways of doing this.

O_{2} : 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.

O_{1} : Department #1 picks delegate and alternate; P_{12, 2}
= 132 ways.

O_{2} : Department #2 picks delegate and alternate; P_{ 15, 2}
= 210 ways.

O_{3} : 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).