# Discrete Math HW Check

1) The board of directors of a pharmaceutical corporation has 10 members. An upcoming stockholder's meeting is scheduled to approve a new slate of company officers (chosen from the 10 board members).

A) 4 (Presendent, Vice Presendent, secretary, and treasurer) positions needs filled. How many possible ways are there to do it?

10 x 9 x 8 x 7

B) Three members of the board of directors are physicians. How many slates from part (A) have a physician nominated for presendency?

3 x 9 x 8 x 7 ????

C) How many slates have exactly one physician?

3 x 7 x 6 x 5

D) How many slates have at least one physician?

3 x 9 x 8 x 7

2) There are 3 cities: a, b, c. City a has two roads that go to C and 4 roads that go to b. City b has 3 roads that go to c.

A) How many ways can you get from a to c?

2 + (4x3)

B) How many round trips from a to c are there such that the return trip is at least partially different than the route taken to get there?

(2 + (4x3)) x (2 + (4x3)) - (2 + (4x3)) ????

3) If n is a positive integer and n is greater than 1, prove that c(n, 2) + c(n-1/2) is a perfect square.

I have no idea

4) A gym coach must select 11 seniors to play on a football team. If he can make his selection in 12,376 ways, how many seniors are eligible to play?

n! / (11! x (n-11)!) = 12,376
n! / (n-11)! = 12,376(11!)
That's all I got

5) How many ways can 10 identical dimes be distributed among five children if:

A) There are no restrictions?

c(14,10)

B) Each child gets at least one dime?

c(9,5)

C) The oldest child gets at least two dimes?

c(12,8)

I know those are the answers, but I don't know why.

6) Determine the number of integer solutions of:
x1 + x2 + x3 + x4 = 32
where:

A) xi >= 0, 1<= i <=4

c(35, 32)

B) xi > 0, 1<= i <=4

c(31, 28) Why?

C) x1, x2 >= 5, x3, x4 >= 7

c(11, 8) Why?

lightgrav
Homework Helper
In 1c, it looks like you're counting
number of Dr.President slates (from B)
that have no other Dr. on it. I would count
yours + (n Dr. n n) + (n n Dr. n) + (n n n Dr.)

In 5, line up the kids and the dimes in a ring, so
each kid gets the dimes clockwise of them.
Now, how many ways can you arrange 15 things
(10 dimes + 3 kids) into the 15 spots on the ring?
If each kid already has been given a dime,
there's 5 dimes left (and 10 spots on the ring).